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Exercises 1 We are told that the equations y=3x−20 and y=-2x+10 intersect at (6,-2). We need to figure out, without solving, the solution to the equation 3x−20=-2x+10. What can we notice about this new equation? Most importantly, it is made of the right-hand sides of the previous equations. Since we know that the point of intersection for the original two equations is (6,-2), we know that -2-2=3⋅6−20 and=-2⋅6+10. Finally, since the left-hand sides are the same, we know that the right-hand sides must also be equal. Therefore, 3⋅6−20=-2=-2⋅6+10, making the solution to our equation x=6. | |

Exercises 2 When the absolute values of two expressions are equal, either the expressions are equal or the opposite of the expressions are equal. For example, if we have the equation, ∣ax+b∣=∣cx+d∣ then we can have either Expressions are equal:Opposite are equal:ax+b=cx+dax+b=-(cx+d). We have the equation ∣2x−4∣=∣-5x+1∣ and asked to write it as two separate systems of equations. First, we can write it as two separate cases as shown above. 2x+4=-5x+1or2x+4=-(-5x+1) We can simplify the second equation a little bit by distributing the negative sign. Then the right-hand side becomes 5x−1. Now, we can allow each side of both equations to be equal to y, then we will have two systems of equations. 2x+4=-5x+1⇒2x+4=-(-5x+1)⇒{y=2x+4y=-5x+1{y=2x+4y=5x−1) | |

Exercises 3 To solve the equation -2x+3=x, we can create two functions out of the left-hand and right-hand sides of the equation. f(x)=-2x+3and g(x)=x The x-coordinate where the graphs of these functions intersect is the solution to our equation. We can identify it from the graph as shown below.The graphs intersect at x=1, which is our solution. We can verify this algebraically by solving the equation. -2x+3=xLHS+2x=RHS+2x3=3xLHS/3=RHS/31=xRearrange equationx=1 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -2x+3=xx=1-2(1)+3=?1a⋅1=a-2+3=?1Add terms1=1 | |

Exercises 4 To solve the equation -3=4x+1, we can create two functions out of the left-hand and right-hand sides of the equation. f(x)=-3and g(x)=4x+1 The x-coordinate where the graphs of these functions intersect is the solution to our equation. We can identify it from the graph as shown below.The graphs intersect at x=-1, which is our solution. We can verify this algebraically by solving the equation. -3=4x+1LHS−1=RHS−1-4=4xLHS/4=RHS/4-1=xRearrange equationx=-1 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -3=4x+1x=-1-3=?4(-1)+1a(-b)=-a⋅b-3=?-4+1Add terms-3=-3 | |

Exercises 5 To solve the equation -x−1=31x+3, we can create two functions out of the left-hand and right-hand sides of the equation. f(x)=-x−1and g(x)=31x+3 The x-coordinate where the graphs of these functions intersect is the solution to our equation. We can identify it from the graph as shown below.The graphs intersect at x=-3, which is our solution. We can verify this algebraically by solving the equation. -x−1=31x+3LHS⋅3=RHS⋅3-3x−3=x+9LHS−x=RHS−x-4x−3=9LHS+3=RHS+3-4x=12LHS/(-4)=RHS/(-4)x=-3 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -x−1=31x+3x=-3-(-3)−1=?31(-3)+3Remove parentheses and change signs3−1=?31(-3)+3a(-b)=-a⋅b3−1=?-1+3Add and subtract terms2=2 | |

Exercises 6 To solve the equation -23x−2=-4x+3, we can create two functions out of the left-hand and right-hand sides of the equation. f(x)=-23x−2and g(x)=-4x+3 The x-coordinate where the graphs of these functions intersect is the solution to our equation. We can identify it from the graph as shown below.The graphs intersect at x=2, which is our solution. We can verify this algebraically by solving the equation. -23x−2=-4x+3LHS⋅2=RHS⋅2-3x−4=-8x+6LHS+8x=RHS+8x5x−4=6LHS+4=RHS+45x=10LHS/5=RHS/5x=2 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -23x−2=-4x+3x=2-23(2)−2=?-4(2)+3(-a)b=-ab-3−2=?-8+3Add and subtract terms-5=-5 | |

Exercises 7 To graph the equation x+4=-x, we create two functions out of the left- and right-hand sides of the equation. f(x)=x+4and g(x)=-x The x-coordinate where the graphs of these functions intersect is the solution to our equation.The graphs intersect at x=-2, which is our solution. We can verify this algebraically by solving the equation. x+4=-xLHS+x=RHS+x2x+4=0LHS−4=RHS−42x=-4LHS/2=RHS/2x=-2 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. x+4=-xx=-2-2+4=?-(-2)Remove parentheses and change signs-2+4=?2Add terms2=2 | |

Exercises 8 To graph the equation 4x=x+3, we create two functions out of the left- and right-hand sides of the equation. f(x)=4xand g(x)=x+3 The x-coordinate where the graphs of these functions intersect is the solution to our equation.The graphs intersect at x=1, which is our solution. We can verify this algebraically by solving the equation. 4x=x+3LHS−x=RHS−x3x=3LHS/3=RHS/3x=1 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. 4x=x+3x=14(1)=?1+3a⋅1=a4=?1+3Add terms4=4 | |

Exercises 9 To graph the equation x+5=-2x−4, we create two functions out of the left- and right-hand sides of the equation. f(x)=x+5and g(x)=-2x−4 The x-coordinate where the graphs of these functions intersect is the solution to our equation.The graphs intersect at x=-3, which is our solution. We can verify this algebraically by solving the equation. x+5=-2x−4LHS+2x=RHS+2x3x+5=-4LHS−5=RHS−53x=-9LHS/3=RHS/3x=-3 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. x+5=-2x−4x=-3-3+5=?-2(-3)−4-a(-b)=a⋅b-3+5=?6−4Add and subtract terms2=2 | |

Exercises 10 To graph the equation -2x+6=5x−1, we create two functions out of the left- and right-hand sides of the equation. f(x)=-2x+6and g(x)=5x−1 The x-coordinate where the graphs of these functions intersect is the solution to our equation.The graphs intersect at x=1, which is our solution. We can verify this algebraically by solving the equation. -2x+6=5x−1LHS+2x=RHS+2x6=7x−1LHS+1=RHS+17=7xLHS/7=RHS/71=xRearrange equationx=1 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -2x+6=5x−1x=1-2(1)+6=?5(1)−1a⋅1=a-2+6=?5−1Add and subtract terms4=4 | |

Exercises 11 To graph the equation 21x−2=9−5x, we create two functions out of the left- and right-hand sides of the equation. f(x)=21x−2and g(x)=9−5x The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. g(x)=9−5xRearrange termsg(x)=-5x+9 Now, let's graph these lines.The graphs intersect at x=2, which is our solution. We can verify this algebraically by solving the equation. 21x−2=9−5xLHS⋅2=RHS⋅2x−4=18−10xLHS+10x=RHS+10x11x−4=18LHS+4=RHS+411x=22LHS/11=RHS/11x=2 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. 21x−2=9−5xx=221(2)−2=?9−5(2)Multiply1−2=?9−10Subtract terms-1=-1 | |

Exercises 12 To graph the equation -5+41x=3x+6, we create two functions out of the left- and right-hand sides of the equation. f(x)=-5+41xand g(x)=3x+6 The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. f(x)=-5+41xRearrange termsf(x)=41x−5 Now, let's graph these lines.The graphs intersect at x=-4, which is our solution. We can verify this algebraically by solving the equation. -5+41x=3x+6LHS⋅4=RHS⋅4-20+x=12x+24LHS−x=RHS−x-20=11x+24LHS−24=RHS−24-44=11xLHS/11=RHS/11-4=xRearrange equationx=-4 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -5+41x=3x+6x=-4-5+41(-4)=?3(-4)+6a(-b)=-a⋅b-5−1=?-12+6Add and subtract terms-6=-6 | |

Exercises 13 To graph the equation 5x−7=2(x+1), we must create two functions out of the left and right-hand sides of the equation. f(x)=5x−7and g(x)=2(x+1) The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. g(x)=2(x+1)Distribute 2g(x)=2x+2 Now, let's graph these lines.The graphs intersect at x=3, which is our solution. We can verify this algebraically by solving the equation. 5x−7=2(x+1)Distribute 25x−7=2x+2LHS−2x=RHS−2x3x−7=2LHS+7=RHS+73x=9LHS/3=RHS/3x=3 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in true statement, our solution is correct. 5x−7=2(x+1)x=35(3)−7=?2(3+1)Multiply15−7=?2(3+1)Add and subtract terms8=?2(4)Multiply8=8 | |

Exercises 14 To graph the equation -6(x+4)=-3x−6, we create two functions out of the left- and right-hand sides of the equation. f(x)=-6(x+4)and g(x)=-3x−6 The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. f(x)=-6(x+4)Distribute -6f(x)=-6x−24 Now, let's graph these lines.The graphs intersect at x=-6, which is our solution. We can verify this algebraically by solving the equation. -6(x+4)=-3x−6Distribute -6-6x−24=-3x−6LHS+3x=RHS+3x-3x−24=-6LHS+24=RHS+24-3x=18LHS/(-3)=RHS/(-3)x=-6 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -6(x+4)=-3x−6x=-6-6(-6+4)=?-3(-6)−6Add terms-6(-2)=?-3(-6)−6-a(-b)=a⋅b12=?18−6Subtract term12=12 | |

Exercises 15 To graph the equation 3x−1=-x+7, we create two functions out of the left and right-hand sides of the equation. f(x)=3x−1and g(x)=-x+7 The x-coordinate where the graphs of these functions intersect is the solution to our equation.The graphs intersect at x=2, which is our one solution for this equation. We can verify this algebraically by solving the equation. 3x−1=-x+7LHS+x=RHS+x4x−1=7LHS+1=RHS+14x=8LHS/4=RHS/4x=2 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. 3x−1=-x+7x=23(2)−1=?-(2)+7Multiply6−1=?-(2)+7Remove parentheses and change signs6−1=?-2+7Add and subtract terms5=5 | |

Exercises 16 To graph the equation 5x−4=5x+1, we create two functions out of the left and right-hand sides of the equation. f(x)=5x−4and g(x)=5x+1 The x-coordinate where the graphs of these functions intersect is the solution to our equation.We can notice that the lines are parallel. Therefore, the graphs do not have a point of intersection. We can verify this algebraically by solving the equation. 5x−4=5x+1LHS−5x=RHS−5x-4≠1 Solving this system of equations resulted in a contradiction, because -4 can never be equal to 1. Therefore, the lines are parallel and do not have a point of intersection. | |

Exercises 17 To graph the equation -4(2−x)=4x−8, we create functions out of the left- and right-hand-side of the equation: f(x)=-4(2−x)and g(x)=4x−8. The x-coordinate where the graphs of these functions intersect, is the solution to our equation.The graphs are the identical which means the functions have infinitely many solutions. | |

Exercises 18 To graph the equation -2x−3=2(x−2), we create two functions out of the left and right-hand sides of the equation. f(x)=-2x−3and g(x)=2(x−2) The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. g(x)=2(x−2)Distribute 2g(x)=2x−4 Now, let's graph these lines.The graphs intersect at x=41, which is our one solution to this equation. We can verify this algebraically by solving the equation. -2x−3=2(x−2)Distribute 2-2x−3=2x−4LHS+2x=RHS+2x-3=4x−4LHS+4=RHS+41=4xLHS/4=RHS/441=xRearrange equationx=41 To check this solution, we will substitute it back into the given equation and simplify. If doing so results in a true statement, our solution is correct. -2x−3=2(x−2)x=41-2(41)−3=?2(41−2)(-a)b=-ab-21−3=?2(41−2)Subtract terms-27=?2(-47)a(-b)=-a⋅b-27=-27 | |

Exercises 19 To graph the equation -x−5=-31(3x+5), we create two functions out of the left and right-hand sides of the equation. f(x)=-x−5and g(x)=-31(3x+5) The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. g(x)=-31(3x+5)Distribute -31g(x)=-x−35 Now let's graph these lines.We can see that the lines are parallel. Therefore, the graphs do not have a point of intersection. We can verify this algebraically by solving the equation. -x−5=-31(3x+5)Distribute -31-x−5=-x−35LHS+x=RHS+x-5≠-35 Solving this system of equations resulted in a contradiction, because -5 can never be equal to -35. Therefore, the lines are parallel and do not have a point of intersection. | |

Exercises 20 To graph the equation 21(8x+3)=4x+23, we create two functions out of the left and right-hand sides of the equation. f(x)=21(8x+3)and g(x)=4x+23 The x-coordinate where the graphs of these functions intersect is the solution to our equation. Since one of the given equations is not in slope-intercept form, let's rewrite it so that it will be easier to identify its slope and y-intercept. f(x)=21(8x+3)Distribute 21f(x)=4x+23 Now, let's graph these lines.We can see that the lines are the same and have infinitely many intersection points. We can verify this algebraically by solving the equation. 21(8x+3)=4x+23Distribute 214x+23=4x+23LHS−4x=RHS−4x23=23 Solving this system of equations resulted in an identity, because 23 is always equal to itself. Therefore, the lines are the same and have infinitely many intersection points. | |

Exercises 21 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣x−4∣=∣3x∣∣x−4∣=∣3x∣x−4∣=-3x We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left- and right-hand sides of the equation. y=x−4and y=3x The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (-2,-6) which means the equation's solution is x=-2. Let's check whether it's correct by substituting it into the original equation. ∣x−4∣=∣3x∣x=-2∣-2−4∣=?∣3(-2)∣ Simplify Subtract term∣-6∣=?∣3(-2)∣a(-b)=-a⋅b∣-6∣=?∣-6∣∣-6∣=6 6=6 The equation is true, so x=-2 is a solution.Second equation In order to graph the second equation, we again create functions out of the left- and right-hand sides of the equation. y=x−4and y=-3x Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (1,-3), which means the solution to this equation is x=1. ∣x−4∣=∣3x∣x=1∣1−4∣=?∣3(1)∣ Simplify Subtract term∣-3∣=?∣3(1)∣a⋅1=a∣-3∣=?∣3∣∣-3∣=33=?∣3∣∣3∣=3 3=3 Checking our solution, we confirmed that x=1 is a solution. | |

Exercises 22 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣2x+4∣=∣x−1∣∣2x+4∣=∣x−1∣2x+4∣=-(x−1) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left- and right-hand sides of the equation. y=2x+4and y=x−1 The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (-5,-6) which means the equation's solution is x=-5. Let's check whether it's correct by substituting it into the original equation. ∣2x+4∣=∣x−1∣x=-5∣2(-5)+4∣=?∣-5−1∣ Simplify a(-b)=-a⋅b∣-10+4∣=?∣-5−1∣Add and subtract terms∣-6∣=?∣-6∣∣-6∣=6 6=6 The equation is true, so x=-5 is a solution.Second equation In order to graph the second equation, we again create functions out of the left- and right-hand sides of the equation. y=2x+4and y=-(x−1) Let's first simplify the form of the second function. y=-(x−1)Remove parentheses and change signsy=-x+1 Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (-1,2), which means the solution to this equation is x=-1. ∣2x+4∣=∣x−1∣x=-1∣2(-1)+4∣=?∣-1−1∣ Simplify a(-b)=-a⋅b∣-2+4∣=?∣-1−1∣Add and subtract terms∣2∣=?∣-2∣∣2∣=22==?∣-2∣∣-2∣=2 2=2 Checking our solution, we confirmed that x=-1 is a solution. | |

Exercises 23 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now, let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣2x∣=∣x+3∣∣2x∣=∣x+3∣2x∣=-(x+3) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left and right-hand sides of the equation. y=2xand y=x+3 The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (3,6) which means the equation's solution is x=3. Let's check whether it's correct by substituting it into the original equation. ∣2x∣=∣x+3∣x=3∣2(3)∣=?∣3+3∣ Simplify Multiply∣6∣=?∣3+3∣Add terms∣6∣=?∣6∣∣6∣=6 6=6 The equation is true, so x=3 is a solution.Second equation In order to graph the second equation, we again create functions out of the left and right-hand sides of the equation. y=2xand y=-(x+3) Let's first simplify the form of the second function. y=-(x+3)Remove parentheses and change signsy=-x−3 As before, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (-1,-2), which means the solution to this equation is x=-1. ∣2x∣=∣x+3∣x=-1∣2(-1)∣=?∣-1+3∣ Simplify a(-b)=-a⋅b∣-2∣=?∣-1+3∣Add terms∣-2∣=?∣2∣∣-2∣=22=?∣2∣∣2∣=2 2=2 By checking our solution, we confirmed that x=-1 is a solution. | |

Exercises 24 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣2x−6∣=∣x∣∣2x−6∣=∣x∣2x−6∣=-x We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left and right-hand sides of the equation. y=2x−6and y=x The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (6,6) which means the equation's solution is x=6. Let's check whether it's correct by substituting it into the original equation. ∣2x−6∣=∣x∣x=6∣2(6)−6∣=?∣6∣ Simplify Multiply∣12−6∣=?∣6∣Subtract term∣6∣=?∣6∣∣6∣=6 6=6 The equation is true, so x=6 is a solution.Second equation In order to graph the second equation, we again create functions out of the left-hand and right-hand sides of the equation. y=2x−6and y=-x Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (2,-2), which means the solution to this equation is x=2. ∣2x−6∣=∣x∣x=2∣2(2)−6∣=?∣2∣ Simplify Multiply∣4−6∣=?∣2∣Subtract term∣-2∣=?∣2∣∣-2∣=22=?∣2∣∣2∣=2 2=2 By checking our solution, we confirmed that x=2 is a solution. | |

Exercises 25 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣-x+4∣=∣2x−2∣∣-x+4∣=∣2x−2∣-x+4∣=-(2x−2) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left-hand and right-hand sides of the equation. y=-x+4and y=2x−2 The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (2,2) which means the equation's solution is x=2. Let's check whether it's correct by substituting it into the original equation. ∣-x+4∣=∣2x−2∣x=2∣-(2)+4∣=?∣2(2)−2∣ Simplify Remove parentheses and change signs∣-2+4∣=?∣2(2)−2∣Multiply∣-2+4∣=?∣4−2∣Add and subtract terms∣2∣=?∣2∣∣2∣=2 2=2 The equation is true, so x=2 is a solution.Second equation In order to graph the second equation, we again create functions out of the left-hand and right-hand sides of the equation. y=-x+4and y=-(2x−2) Let's first simplify the form of the second function. y=-(2x−2)Remove parentheses and change signsy=-2x+2 Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (-2,6), which means the solution to this equation is x=-2. ∣-x+4∣=∣2x−2∣x=-2∣-(-2)+4∣=?∣2(-2)−2∣ Simplify Remove parentheses and change signs∣2+4∣=?∣2(-2)−2∣a(-b)=-a⋅b∣2+4∣=?∣-4−2∣Add and subtract terms∣6∣=?∣-6∣∣6∣=66=?∣-6∣∣-6∣=6 6=6 By checking our solution, we confirmed that x=-2 is a solution. | |

Exercises 26 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣x+2∣=∣-3x+6∣∣x+2∣=∣-3x+6∣x+2∣=-(-3x+6) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left-hand and right-hand sides of the equation. y=x+2and y=-3x+6 The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (1,3) which means the equation's solution is x=1. Let's check whether it's correct by substituting it into the original equation. ∣x+2∣=∣-3x+6∣x=1∣1+2∣=?∣-3(1)+6∣ Simplify a⋅1=a∣1+2∣=?∣-3+6∣Add terms∣3∣=?∣3∣∣3∣=3 3=3 The equation is true, so x=1 is a solution.Second equation In order to graph the second equation, we again create functions out of the left and right-hand sides of the equation. y=x+2and y=-(-3x+6) Let's first simplify the form of the second function. y=-(-3x+6)Remove parentheses and change signsy=3x−6 Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (4,6), which means the solution to this equation is x=4. ∣x+2∣=∣-3x+6∣x=4∣4+2∣=?∣-3(4)+6∣ Simplify (-a)b=-ab∣4+2∣=?∣-12+6∣Add and subtract terms∣6∣=?∣-6∣∣6∣=66=?∣-6∣∣-6∣=6 6=6 By checking our solution, we confirmed that x=4 is a solution. | |

Exercises 27 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣x+1∣=∣x−5∣∣x+1∣=∣x−5∣x+1∣=-(x−5) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left-hand and right-hand sides of the equation. y=x+1and y=x−5 The x-coordinate where the graphs of these functions intersect is the solution to our equation.We have found that these lines are parallel, and therefore will never have a point of intersection. This means that for this possible configuration of the original equation, there is no solution.Second equation In order to graph the second equation, we again create functions out of the left-hand and right-hand sides of the equation. y=x+1and y=-(x−5) Let's first simplify the form of the second function. y=-(x−5)Remove parentheses and change signsy=-x+5 Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (2,3), which means the solution to this equation is x=2. ∣x+1∣=∣x−5∣x=2∣2+1∣=?∣2−5∣ Simplify Add and subtract terms∣3∣=?∣-3∣∣3∣=33=?∣-3∣∣-3∣=3 3=3 By checking our solution, we confirmed that x=2 is a solution. | |

Exercises 28 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣2x+5∣=∣-2x+1∣∣2x+5∣=∣-2x+1∣2x+5∣=-(-2x+1) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left-hand and right-hand sides of the equation. y=2x+5and y=-2x+1 The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (-1,3) which means the equation's solution is x=-1. Let's check whether it's correct by substituting it into the original equation. ∣2x+5∣=∣-2x+1∣x=-1∣2(-1)+5∣=?∣-2(-1)+1∣ Simplify a(-b)=-a⋅b∣-2+5∣=?∣-2(-1)+1∣-a(-b)=a⋅b∣-2+5∣=?∣2+1∣Add and subtract terms∣3∣=?∣3∣∣3∣=3 3=3 The equation is true, so x=-1 is a solution.Second equation In order to graph the second equation, we again create functions out of the left and right-hand sides of the equation. y=2x+5and y=-(-2x+1) Let's first simplify the form of the second function. y=-(-2x+1)Remove parentheses and change signsy=2x−1 Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.We have found that these lines are parallel, and therefore will never have a point of intersection. This means that for this possible configuration of the original equation, there is no solution. | |

Exercises 29 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. First, notice that, using the properties of absolute value, we can rewrite the equation as shown below. ∣x−3∣=2∣x∣⇔∣x−3∣=∣2x∣ Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:∣x−3∣=2∣x∣∣x−3∣=∣2x∣x−3∣=-2x We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left-hand and right-hand sides of the equation. y=x−3and y=2x The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (-3,-6) which means the equation's solution is x=-3. Let's check whether it's correct by substituting it into the original equation. ∣x−3∣=2∣x∣x=-3∣-3−3∣=?2∣-3∣ Simplify Subtract term∣-6∣=?2∣-3∣∣-6∣=66=?2∣-3∣∣-3∣=36==?2(3)Multiply 6=6 The equation is true, so x=-3 is a solution.Second equation In order to graph the second equation, we again create functions out of the left-hand and right-hand sides of the equation. y=x−3and y=-2x Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (1,-2), which means the solution to this equation is x=1. ∣x−3∣=2∣x∣x=1∣1−3∣=?2∣1∣ Simplify Subtract term∣-2∣=?2∣1∣∣1∣=1-2∣=?2(1)a⋅1=a∣-2∣=?2∣-2∣=2 2=2 By checking our solution, we confirmed that x=1 is a solution. | |

Exercises 30 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d) Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. First, notice that, using the properties of absolute value and distributive property, we can rewrite the equation as shown below. 4∣x+2∣=∣2x+7∣⇔∣4x+8∣=∣2x+7∣ Now let's consider these two cases for the given equation. Given equation:First equation:Second equation:4∣x+2∣=∣2x+7∣∣4x+8∣=∣2x+7∣4x+8∣=-(2x+7) We'll solve each of these equations by graphing separately.First equation To graph the first equation, we create two functions out of the left-hand and right-hand sides of the equation. y=4x+8and y=2x+7 The x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the graphs intersect at (-0.5,-6) which means the equation's solution is x=-0.5. Let's check whether it's correct by substituting it into the original equation. 4∣x+2∣=∣2x+7∣x=-0.54∣-0.5+2∣=?∣2(-0.5)+7∣ Simplify a(-b)=-a⋅b4∣-0.5+2∣=?∣-1+7∣Add terms4∣1.5∣=?∣6∣∣1.5∣=1.54((1.5)=?∣6∣Multiply6=?∣6∣∣6∣=6 6=6 The equation is true, so x=-0.5 is a solution.Second equation In order to graph the second equation, we again create functions out of the left-hand and right-hand sides of the equation. y=4x+8and y=-(2x+7) Let's first simplify the form of the second function. y=-(2x+7)Remove parentheses and change signsy=-2x−7 Once more, the x-coordinate where the graphs of these functions intersect is the solution to our equation.From the graph, we can see that the lines intersect at (-2.5,-2), which means the solution to this equation is x=-2.5. 4∣x+2∣=∣2x+7∣x=-2.54∣-2.5+2∣=?∣2(-2.5)+7∣ Simplify a(-b)=-a⋅b4∣-2.5+2∣=?∣-5+7∣Add terms4∣-0.5∣=?∣2∣∣-0.5∣=0.54(0.5)=?∣2∣Multiply2=?∣2∣∣2∣=2 2=2 By checking our solution, we confirmed that x=-2.5 is a solution. | |

Exercises 31 To graph the equation, we first have to rewrite the left-hand side and right-hand side as linear functions in slope-intercept form: y=0.7x+0.5andy=-0.2x−1.3 Let's enter these equations in our calculator by pressing Y= and then writing them on the first two rows.When the equations have been entered, we can plot them by pressing GRAPH.To find the intersection point, we can use the tool "intersect", which we find by pressing 2nd and then TRACE.Having selected intersect, you will see the plot again. Now you have to choose the graphs between which you want to calculate the intersection point.First curve: Choose the first graph by pressing ENTER. Second curve: Choose the second graph by pressing ENTER once more. Guess: In order for the calculator to determine the intersection point, it will ask you for a guess which will make the calculation go faster. Place the marker close to the intersection point and then press enter. The calculator will now display the intersection of the curves including it's x- and y-value. | |

Exercises 32 To graph the equation, we first have to rewrite the left-hand side and right-hand side as linear functions in slope-intercept form: y=2.1x+0.6andy=-1.4x+6.9 Let's enter these equations in our calculator by pressing Y= and then writing them on the first two rows.When the equations have been entered, we can plot them by pressing GRAPH.To find the intersection point, we can use the tool "intersect", which we find by pressing 2nd and then TRACE.Having selected intersect, you will see the plot again. Now you have to choose the graphs between which you want to calculate the intersection point.First curve: Choose the first graph by pressing ENTER. Second curve: Choose the second graph by pressing ENTER once more. Guess: In order for the calculator to determine the intersection point, it will ask you for a guess which will make the calculation go faster. Place the marker close to the intersection point and then press enter. The calculator will now display the intersection of the curves including it's x- and y-value. | |

Exercises 33 | |

Exercises 34 To begin, we will create two equations. One to represent the age of the dog (in dog years) and one to represent the age of the cat (in cat years).Dog. It is given that the current age of the dog in dog years is 16. Also, for every human year, the dog ages 7 dog years. If we let x represent the number of human years that have passed and y represent the age of the dog, we can write the following equation for the dogs age in dog years:y=16+7x Cat. It is given that the current age of the cat in cat years is 28. Also, for every human year, the cat ages 4 cat years. Since x represent the number of human years that have passed and y represent the age of the cat, we can write the following equation for the cats age in cat years:y=28+4xTo find the number of human years it will take for the animals to reach the same age in their respective years, means to find the value of x that makes the equations equal the same y value. We can determine this graphically.From the graph, we can see that the lines intersect at (4,44). This means that after 4 more human years, both animals will be 44 years old in their respective years. | |

Exercises 35 | |

Exercises 36 If we consider the equations as a system of equations, it would be true that the solution to the system is (4,0). This is because the solution to a system is the (x,y) point that solves both equations. However, (4,0) is not the solution to the equation. Notice that when we set the equations equal to each other, there is only one variable left, x. This is because setting the equations equal to each other implies that the y values are equal. The solution to -x+4=2x−8 will therefore be the x-coordinate of the point where y=-x+4andy=2x−8 intersect. Since (4,0) is the intersection point the solution is x=4. | |

Exercises 37 There are infinitely many solutions to this exercise. What follows is one possible example. To begin, if x=-3 is in fact a solution to the equation it means we can substitute x=-3 into the equation and it will make a true statement. To determine the necessary values of m and b, we will hence begin by substituting x=-3 into the equation and simplify mx+b=-2x−1x=-3m(-3)+b=-2(-3)−1Multiply-3m+b=6−1Subtract terms-3m+b=5 At this point, our equation is -3m+b=5. We can now arbitrarily choose the value of one of the variables. For example, if we let m=1, we can find the corresponding value of b. -3m+b=5m=1-3⋅1+b=5a⋅1=a-3+b=5LHS+3=RHS+3b=8 Thus, if m=1 then b=8 so our equation becomes x+8=-2x−1. By graphing the lines y=x+8 and y=-2x−1 we can make sure that the x-coordinate of the intersection point is x=-3.The lines intersect at (-3,5) which means the solution to the equation is in fact x=-3. | |

Exercises 38 aTo estimate the point of intersection, we can note the approximate x- and y- coordinates of the point where the lines cross. From the graph, we can see that the point of intersection is approximately (7,5.5).bIt is given that x represents the number of years the company has been in business and y represents the total revenue of the company in billions of dollars. In Part A, we found that the solution to the system was approximately (7,5.5) That means x=7 and y=5.5. In context, we can say that 7 years after the company has been in business, their total revenue is 5.5 million dollars. | |

Exercises 39 | |

Exercises 40 We have to look at each car separately.Original Car It is given that the original car has an initial value of 20000 dollars and the value decreases by 1500 dollars each year. If we let x represent the number of years that passes and y represent the value of the car, we can write the following equation: y=20000−1500x. To determine the value of the original car after 5 years, we can substitute x=5 into the equation and solve for y. y=20000−1500xx=5y=20000−1500⋅5Multiplyy=20000−7500Subtract termsy=12500 The value of the car at year 5 will be 12500 dollar.Different Car For the different car to have the same value as the original car at 5 years, it must also be worth 12500. We can use the equation y=mx+b to represent the value of the different car, where b is the initial value and m is the rate per year by which the value decreases. We will assume that (5,12500) is a solution to the equation. That will ensure this car meets the requirements of the exercise. So let's begin by substituting (5,12500) into the equation for (x,y). y=mx+bx=5, y=1250012500=m⋅5+bMultiply12500=5m+b Now we can arbitrarily choose the initial value of the car. Suppose this car was initially worth 18000 so we have b=18000. By substituting this value into our equation we can solve for m. 12500=5m+bb=1800012500=5m+18000 Solve for m LHS−18000=RHS−18000-5500=5mLHS/5=RHS/5-1100=mRearrange equation m=-1100 If this car is initially worth 18000, it must decrease by 1100 dollars each year, to reach 12500 after 5 years. We can now write the equation as y=-1100x+18000. We can graph both equations to ensure that both cars are worth 12500 after 5 years.Since both lines pass through the point (5,12500), we can conclude that both cars are worth 12500 dollars in year 5. | |

Exercises 41 aThe solution of the equation ax+b=cx+d will be the x-coordinate of the point where the lines y=ax+b and y=cx+d intersect. To simplify this solution we will analyze the lines separately. Notice that both of these equations are given in the form y=mx+b, where m is the slope of the line and b is the y-intercept. Thus, a and c represent the slopes of the lines while b and d represent the y-intercepts. It is given that 0<b<d and a<c. We will begin by analyzing the y-intercepts. Since both b and d are greater than 0, both y-intercepts are positive, and thus lie above the x-axis. Additionally, since b<d, d lies further up the y-axis than b. We can conceptually sketch the graphs y-intercepts as follows.Now that we have placed the y-intercepts, we can sketch the lines using the slopes. Since a<c, the slope of y=cx+d will be steeper than the slope of y=ax+b.From the graph, we can see that the lines intersect at a negative x value. Thus, the solution to ax+b=cx+d is negative.bIt is given that d<b<0 and a<c. We will begin by analyzing the y-intercepts. Since both b and d are less than 0, both y-intercepts are negative, and thus lie below the x-axis. Additionally, since d<b, d lies further down the y-axis than b. We can conceptually sketch the graphs y-intercepts as follow.Now that we have placed the y-intercepts, we can sketch the lines using the slopes. Since a<c, the slope of y=cx+d will be steeper than the slope of y=ax+b. We can now conceptually draw the graphs.From the graph, we can see that the lines intersect at a positive x value. Thus, the solution to ax+b=cx+d is positive. | |

Exercises 42 To graph an inequality on a number line, we need to make note of two things: the direction of the inequality and whether it is strict. y>5 Observing the given inequality, we see that all values of y are greater than 5. This means that the solution set lies to the right of 5 on a number line. Furthermore, the inequality symbol indicates that y=5 is not a solution, which we will mark with an open circle on the number line at 5. | |

Exercises 43 To graph an inequality on a number line, we need to make note of two things: the direction of the inequality and whether it is strict. x≤-2 Observing the given inequality, we see that all values of x are less than or equal to -2. This means that the solution set lies to the left of -2 on a number line. Furthermore, the inequality symbol indicates that x=-2 is a solution, which we will mark with a closed circle on the number line at -2. | |

Exercises 44 To graph an inequality on a number line, we need to make note of two things: the direction of the inequality and whether it is strict. n≥9 Observing the given inequality, we see that all values of n are greater than or equal to 9. This means that the solution set lies to the right of 9 on a number line. Furthermore, the inequality symbol indicates that n=9 is a solution, which we will mark with a closed circle on the number line at 9. | |

Exercises 45 To graph an inequality on a number line, we need to make note of two things: the direction of the inequality and whether it is strict. c<-6 Observing the given inequality, we see that all values of c are less than -6. This means that the solution set lies to the left of -6 on a number line. Furthermore, the inequality symbol indicates that c=-6 is not a solution, which we will mark with an open circle on the number line at -6. | |

Exercises 46 Let's graph f(x) first, then we can graph g(x) on the same coordinate system to compare them.Graphing f(x) We can make a table of values to find points of the graph of f(x)=x−5.xx−5f(x) -2-2−5-7 00−5-5 22−5-3 Let's plot these points and connect them with a straight line to obtain the graph of f(x).Graphing g(x) Now, let's look at how the function g(x)=f(x+2) differs from f(x).xf(x)f(x+2)g(x) -2-7-2+2−5-5 0-50+2−5-3 2-32+2−5-1 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 2 units to the left. This is a horizontal translation. | |

Exercises 47 Let's graph f(x) first, then can graph g(x) on the same coordinate system to compare.Graphing f(x) We can make a table of values to find points of the graph of f(x)=6x.x6xf(x) -26(-2)-12 06(0)0 26(2)12 Let's plot these points and connect them with a straight line to obtain the graph of f(x).Graphing g(x) Now, let's look at how the function g(x)=-f(x) differs from f(x).xf(x)-f(x)g(x) -2-12-(-12)12 00-(0)0 212-(12)-12 We can see that the only thing that has changed is the sign of the y-coordinate.The points are at the same distance from the x-axis. We call this a reflection in the x-axis. | |

Exercises 48 Let's graph f(x) first, then we can graph g(x) on the same coordinate system to compare.Graphing f(x) We can make a table of values to find points for the graph of f(x)=-2x+1.x-2x+1f(x) -1-2(-1)+13 0-2(0)+11 1-2(1)+1-1 Let's plot these points and connect them with a straight line to obtain the graph of f(x).Graphing g(x) Now, let's look at how the function g(x)=f(4x) differs from f(x).xf(x)f(4x)g(x) -13-2(4(-1))+19 01-2(4(0))+11 1-1-2(4(1))+1-7 If we plot these points on the same coordinate plane as f(x), we can see that our points have shrunk four times as close to the y-axis as they were before.This is called a horizontal shrink by a factor of 41. | |

Exercises 49 Let's graph f(x) first, then we can graph g(x) on the same coordinate system to compare.Graphing f(x) We can make a table of values to find points for the graph of f(x)=21x−2.x21x−2f(x) -221(-2)−2-3 121(1)−2-23 421(4)−20 Let's plot these points and connect them with a straight line to obtain the graph of f(x).Graphing g(x) Now, let's look at how the function g(x)=f(x−1) differs from f(x).xf(x)f(x−1)g(x) -2-321(-2−1)−2-27 1-2321(1−1)−2-2 4021(4−1)−2-21 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 1 unit to the right. This is a horizontal translation. |

##### Other subchapters in Solving Systems of Linear Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Solving Systems of Linear Equations by Graphing
- Solving Systems of Linear Equations by Substitution
- Solving Systems of Linear Equations by Elimination
- Solving Special Systems of Linear Equations
- Quiz
- Graphing Linear Inequalities in Two Variables
- Systems of Linear Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment