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Exercises 1 A system of linear equations is simply two or more linear equations containing the same variables. The given equations are both linear and both contain x and y variables. Therefore, they form a system of equations. 5y−2x=186x=-4y−10 | |

Exercises 2 Let's look at each "question" individually and then compare our answers.Solve the system of linear equations. We are first asked to solve the system of linear equations. We can do this by solving each equation for y, graphing them both, and finding their point of intersection. The given equations are: -4x+2y=44x−y=-6 Let's solve the first equation for y so that we may easily graph it using the slope-intercept form. -4x+2y=4LHS+4x=RHS+4x2y=4x+4LHS/2=RHS/2y=2x+2 Now, we can solve the second equation. 4x−y=-6 Solve for y LHS+y=RHS+y4x=-6+yLHS+6=RHS+64x+6=yRearrange equation y=4x+6 Next, we can use the slope-intercept form to graph the equations and find their point of intersection.This graph shows us that the point (-2,-2) lies on both lines and solves the system of equations.Solve each equation for y. To solve each equation for y, we simply need to rewrite each equation into slope-intercept form using a series of inverse operations. Let's begin with the first equation. -4x+2y=4 Solve for y LHS+4x=RHS+4x2y=4x+4LHS/2=RHS/2y=24x+4Factor out 2y=22(2x+2)ba=b/2a/2 y=2x+2 Now we can do the same thing with the second equation. 4x−y=-6 Solve for y LHS+y=RHS+y4x=-6+yLHS+6=RHS+64x+6=yRearrange equation y=4x+6 When we solve each equation for y, we get: y=2x+2y=4x+6.Find the point of intersection. To find the point of intersection of the graphs of the equations, we need to first rewrite the equations into slope-intercept form. As we have previously shown, the equations in slope-intercept form are: -4x+2y=44x−y=-6⇒y=2x+2⇒y=4x+6 We then graph those lines and find where they intersect.The point of intersection is (-2,-2).Find an ordered pair that is a solution. To find an ordered pair that is a solution of each equation in the system, we need to graph the lines and find the point that lies on both lies, their point of intersection. As shown previously, the lines written in slope-intercept form are: -4x+2y=44x−y=-6⇒y=2x+2⇒y=4x+6 We then graph those lines and find a point that lies on both lines.The only ordered pair that lies on both lines is (-2,-2), which is also their point of intersection.Conclusion As shown above, being asked to solve a system of equations, find the point of intersection, or find an ordered pair that solves both equations,are all the same thing. The point of intersection is the only point that lies on both lines and it, subsequently, solves the system of equations. Being asked to solve for y in each equation is it's own, separate question. However, it is often a vital step in the process of solving the system of equations. | |

Exercises 3 To tell whether the point is a solution of the system of linear equations, we can substitute it into each of the equations and simplify. If they hold true, it's a solution.First equation Let's substitute the point into the first equation. x+y=8x=2, y=62+6=?8Add terms8=8 The point is a solution to the first equation because 8 equals 8.Second equation Now we can do the same for the second equation. 3x−y=0x=2, y=63⋅2−(6)=?0 Simplify LHS Remove parentheses and change signs3⋅2−6=?0Multiply6−6=?0Subtract term 0=0 Since 0 is equal to 0, the point is a solution to this equation of the system as well. Since it satisfies both equations, the point is a solution to the system. | |

Exercises 4 To tell whether the point is a solution of the system of linear equations, we can simply substitute it into each of the equations and simplify. If they hold true, it's a solution.First equation Let's substitute the point into the equation. x−y=6x=8, y=28−2=?6Subtract term6=6 The point is a solution to the first equation because 6 equals 6.Second equation Now we can do the same for the second equation. 2x−10y=4x=8, y=22⋅8−10⋅2=?4Multiply16−20=?4Subtract term-4=4 Since -4 is not equal to 4, the statement isn't true after substituting the point. This means the point is not a solution to this equation of the system. | |

Exercises 5 When determining if a given point satisfies just one equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of equations, we apply the same method to all equations in the system. The difference, however, is that if even one substitution results in a false statement, then the point cannot be a solution to the system. Let's substitute (-1,3) into both equations in the given system and find out if it is a solution. {y=-7x−4y=8x+5(I)(II)x=-1, y=3{3=?-7(-1)−43=?8(-1)+5(I): -a(-b)=a⋅b{3=?7−43=?8(-1)+5(II): a(-b)=-a⋅b{3=?7−43=?-8+5Add and subtract terms{3=33=-3 One of our resulting statements, 3=-3, is not true. The point (-1,3) satisfies (I) but does not satisfy (II). Because both statements are not true, the point (-1,3) is not contained in the solution set of the system. | |

Exercises 6 When determining if a given point satisfies just one equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of equations, we apply the same method to all equations in the system. The difference, however, is that if even one substitution results in a false statement, then the point cannot be a solution to the system. Let's substitute (-4,-2) into both equations in the given system and find out if it is a solution. {y=2x+6y=-3x−14(I)(II)x=-4, y=-2{-2=?2(-4)+6-2=?-3(-4)−14(I): a(-b)=-a⋅b{-2=?-8+6-2=?-3(-4)−14(II): -a(-b)=a⋅b{-2=?-8+6-2=?12−14Add and subtract terms{-2=-2-2=-2 Since -2 is equal to -2, both statements are true. Therefore, the point (-4,-2) is contained in the solution set of the system. | |

Exercises 7 When determining if a given point satisfies just one equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of equations, we apply the same method to all equations in the system. The difference, however, is that if even one substitution results in a false statement, then the point cannot be a solution to the system. Let's substitute (-2,1) into both equations in the given system and find out if it is a solution. {6x+5y=-72x−4y=-8(I)(II)x=-2, y=1{6(-2)+5(1)=?-72(-2)−4(1)=?-8(I), (II): a⋅1=a{6(-2)+5=?-72(-2)−4=?-8(I), (II): a(-b)=-a⋅b{-12+5=?-7-4−4=?-8Add and subtract terms{-7=-7-8=-8 Since -7 is equal to -7 and -8 is equal to -8, both statements are true. Therefore, the point (-2,1) is contained in the solution set of the system. | |

Exercises 8 When determining if a given point satisfies just one equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set of the equation. For systems of equations, we apply the same method to all equations in the system. The difference, however, is that if even one substitution results in a false statement, then the point cannot be a solution to the system. Let's substitute (5,-6) into both equations in the given system and find out if it is a solution. {6x+3y=124x+y=14(I)(II)x=5, y=-6{6(5)+3(-6)=?124(5)+(-6)=?14(I), (II): Multiply{30+3(-6)=?1220+(-6)=?14(I): a(-b)=-a⋅b{30−18=?1220+(-6)=?14(II): Remove parentheses{30−18=?1220−6=?14Add and subtract terms{12=1214=14 Since 12 is equal to 12 and 14 is equal to 14, both statements are true. Therefore, the point (5,-6) is contained in the solution set of the system. | |

Exercises 9 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the graph, we can tell that this point is (1,-3).Let's check this solution by substituting the point into both equations. {x−y=44x+y=1(I)(II)(I), (II): x=1, y=-3{1−(-3)=?44(1)+(-3)=?1(I): a−(-b)=a+b{1+3=?44(1)+(-3)=?1(I): Add terms{4=44(1)+(-3)=?1(II): a⋅1=a{4=44+(-3)=?1(II): a+(-b)=a−b{4=44−3=?1(II): Subtract term{4=41=1 Since both equations hold true after substituting the point, we know that it is the solution of the system of equations. | |

Exercises 10 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the graph, we can tell that this point is (3,2).Let's check this solution by substituting the point into each equation.First Equation Let's substitute (3,2) into the first equation and simplify. x+y=5x=3, y=23+2=?5Add terms5=5 The equation is true.Second Equation Let's now do the same with the second equation. y−2x=-4x=3, y=22−2(3)=?-4Multiply2−6=?-4Subtract term-4=-4 The second equation is also true.Conclusion Since both equations hold true after substituting the point, it is a solution of the system of equations. | |

Exercises 11 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the graph, we can tell that this point is (-4,5).Let's check this solution by substituting the point into each equation.First Equation Let's substitute (-4,5) into the first equation and simplify. 6y+3x=18x=-4, y=56(5)+3(-4)=?18Multiply30+3(-4)=?18a(-b)=-a⋅b30−12=?18Subtract term18=18 The equation is true.Second Equation Let's now do the same with the second equation. -x+4y=24x=-4, y=5-(-4)+4(5)=?24Remove parentheses and change signs4+4(5)=?24Multiply4+20=?24Add terms24=24 The second equation is also true.Conclusion Since both equations hold true after substituting the point, it is a solution of the system of equations. | |

Exercises 12 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the graph, we can tell that this point is (1,-3).Let's check this solution by substituting the point into each equation.First Equation Let's substitute (0,2) into the first equation and simplify. 2x−y=-2x=0, y=22(0)−2=?-2Multiply0−2=?-2Subtract term-2=-2 The equation is true.Second Equation Let's now do the same with the second equation. 2x+4y=8x=0, y=22(0)+4(2)=?8Multiply0+8=?8Add terms8=8 The second equation is also true.Conclusion Since both equations hold true after substituting the point, it is a solution of the system of equations. | |

Exercises 13 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Writing in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the values of m and b.Given EquationSlope-Intercept FormSlope my-intercept b y=-x+7y=-1x+7-17 y=x+1y=1x+111 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a straight edge.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (3,4). This is the solution to the system. | |

Exercises 14 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Write in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b y=-x+4y=-1x+4-14 y=2x−8y=2x+(-8)2-8 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a line.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (4,0). This is the solution to the system. | |

Exercises 15 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Write in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b y=31x+2y=31x+2312 y=32x+5y=32x+5325 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a line.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (-9,-1). This is the solution to the system. | |

Exercises 16 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Write in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b y=43x−4y=43x+(-4)43-4 y=-21x+11y=-21x+11-2111 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a line.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (12,5). This is the solution to the system. | |

Exercises 17 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Write in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b 9x+3y=-3y=-3x+(-1)-3-1 2x−y=-4y=2x+424 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a line.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (-1,2). This is the solution to the system. | |

Exercises 18 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Write in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b 4x−4y=20y=1x+(-5)1-5 y=-5y=0x+(-5)0-5 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a line.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (0,-5). This is the solution to the system. | |

Exercises 19 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Write in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b x−4y=-4y=41x+1411 -3x−4y=12y=-43x+(-3)-43-3 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a line.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (-4,0). This is the solution to the system. | |

Exercises 20 By graphing the given equations, we can determine the solution to the system. This will be the point at which the lines intersect. To do this, we will need the equations to be in slope-intercept form to help us identify the slope m and y-intercept b.Writing in Slope-Intercept Form Let's rewrite each of the equations in the system in slope-intercept form, highlighting the values of m and b.Given EquationSlope-Intercept FormSlope my-intercept b 3y+4x=3y=-34x+1-341 x+3y=-6y=-31x+(-2)-31-2 Graphing the System To graph these equations, we will start by plotting their y-intercepts. Then, we will use the slope to determine another point that satisfies each equation, and connect the points with a straight edge.We can see that the lines intersect at exactly one point.It appears that the lines intersect at (3,-3). This is the solution to the system. | |

Exercises 21 Let's check whether the equations were graphed correctly by rewriting both of them in slope-intercept form first. {x−3y=62x−3y=3(I)(II) Write in slope-intercept form (I), (II): LHS+3y=RHS+3y{x=6+3y2x=3+3y(I): LHS−6=RHS−6{x−6=3y2x=3+3y(II): LHS−3=RHS−3{x−6=3y2x−3=3y(I), (II): LHS/3=RHS/3{31x−2=y32x−1=y(I), (II): Rearrange equation {y=31x−2y=32x−1 Now let's analyze the two graphs by finding their y-intercepts and slope. Then we can check whether the graphed lines match our equations in slope-intercept form.We can see that the equations of the lines are {y=31x−2y=32x−3, This means that 2x−3y=3 wasn't graphed correctly. Let's correct this error by graphing the system of equations that we found previously: {y=31x−2y=32x−1 and determining the point of intersection of the lines.The solution of the system is (-3,-3). | |

Exercises 22 Let's take a close look to the given equations, identifying their slope m and y-intercept b. Equation (I)y=2x+(-1)Equation (II)y=1x+1 To graph them, we will plot the y-intercepts and obtain a second point on each line using the slopes. Then, we will connect the points with a straight edge.We can see above that the lines intersect when x=2. However, when stating the solution to a system of equations, we must state the values of both x and y. The correct answer is (2,3). | |

Exercises 23 You can solve system of equations by using a graphing calculator. Before we can do that, we have to write both equations in Slope-Intercept Form. {0.2x+0.4y=4-0.6x+0.6y=-3(I)(II) (I), (II): Write in slope-intercept form (I): LHS−0.2x=RHS−0.2x{0.4y=-0.2x+4-0.6x+0.6y=-3(I): LHS/0.4=RHS/0.4{y=-0.5x+10-0.6x+0.6y=-3(II): LHS+0.6x=RHS+0.6x{y=-0.5x+100.6y=0.6x−3(II): LHS/0.6=RHS/0.6 {y=-0.5x+10y=x−5 Now we can enter the equations in the calculator. To do that, we press the button Y= and then write the equations on the two first rows.Having entered the equations, you can plot them by pressing GRAPH.Woops! To see the point of intersection we need to increase the window setting. We can do that by pressing the button WINDOW on our graphing calculator.Next, to find the point of intersection, we press 2nd and TRACE and choose the fifth option in the menu, intersect.You will see the graph again. Now you have to select the first and second curve before guessing where the point of intersection is. Make sure you place the cursor as close as possible to the point of intersection. | |

Exercises 24 You can solve system of equations by using a graphing calculator. Before we can do that, we have to write both equations in slope-intercept form. {-1.6x−3.2y=-242.6x+2.6y=26(I)(II) (I), (II): Write in slope-intercept form (I): LHS⋅(-1)=RHS⋅(-1){1.6x+3.2y=242.6x+2.6y=26(I): LHS−1.6x=RHS−1.6x{3.2y=-1.6x+242.6x+2.6y=26(I): LHS/3.2=RHS/3.2{y=-0.5x+7.52.6x+2.6y=26(II): LHS−2.6x=RHS−2.6x{y=-0.5x+7.52.6y=-2.6x+26(II): LHS/2.6=RHS/2.6 {y=-0.5x+7.5y=-x+10 Now we can enter the equations in the calculator. To do that, we press the button Y= and then write the equations on the two first rows.Having entered the equations, you can plot them by pressing GRAPH.Next, to find the point of intersection, we press CALC (2nd + TRACE) and choose the fifth option in the menu, intersect.You will see the graph again. Now you have to select the first and second curve before guessing where the point of intersection is. Make sure you place the cursor as close as possible to the point of intersection.The solution to the system of equations is {x=5y=5. | |

Exercises 25 You can solve system of equations by using a graphing calculator. Before we can do that, we have to write both equations in slope-intercept form. {-7x+6y=00.5x+y=2(I)(II) (I), (II): Write in slope-intercept form (I): LHS+7x=RHS+7x{6y=7x0.5x+y=2(I): LHS/6=RHS/6{y=67x0.5x+y=2(II): LHS−0.5x=RHS−0.5x {y=67xy=-0.5x+2 Now we can enter the equations in the calculator. To do that, we press the button Y= and then write the equations on the two first rows.Having entered the equations, you can plot them by pressing GRAPH.Next, to find the point of intersection, we press CALC (2nd + TRACE) and choose the fifth option in the menu, intersect.You will see the graph again. Now you have to select the first and second curve before guessing where the point of intersection is. Make sure you place the cursor as close as possible to the point of intersection.The solution to the system of equations is {x=1.2y=1.4. | |

Exercises 26 You can solve systems of equations by using a graphing calculator. Before we can do that, we have to write both equations in slope-intercept form. {4x−y=1.52x+y=1.5(I)(II) (I), (II): Write in slope-intercept form (I): LHS−4x=RHS−4x{-y=-4x+1.52x+y=1.5(I): LHS⋅(-1)=RHS⋅(-1){y=4x−1.52x+y=1.5(II): LHS−2x=RHS−2x {y=4x−1.5y=-2x+1.5 Now we can enter the equations in the calculator. To do that, we press the button Y= and then write the equations on the two first rows.Having entered the equations, you can plot them by pressing GRAPH.Next, to find the point of intersection, we press CALC (2nd + TRACE) and choose the fifth option in the menu, intersect.You will see the graph again. Now you have to select the first and second curve before guessing where the point of intersection is. Make sure you place the cursor as close as possible to the point of intersection.The solution to the system of equations is {x=0.5y=0.5. | |

Exercises 27 We are told that you have total of 40 minutes to exercise. If we let t be the time you exercise on the elliptical trainer and b be the time you exercise on stationary bike, we can write an equation to represent the situation. 40=t+b We can rearrange this equation, so it will be easier to plot it later. b=-t+40 We are also told that you burn 8 calories per minute on elliptical trainer and 6 calories per minute on stationary bike. With that information we can calculate how much calories you will burn in total.Verbal expressionAlgebraic expression Calories burnt on elliptical trainer per minute8 Time spent on elliptical trainert Total calories burnt on elliptical trainer8t Calories burnt on stationary bike per minute6 Time spent on stationary bikeb Total calories burnt on stationary bike6b Total calories burnt on both machines8t+6bLet's build an equation that compare total calories burnt with given goal of 300 calories. 300=8t+6b Isolating b will help us graph this equation. 300=8t+6bLHS−8t=RHS−8t-8t+300=6bLHS/6=RHS/6-34t+50=bRearrange equationb=-34t+50 Now that we have both equations in the slope-intercept form we can plot them and find the point of interception.From the graph we can see that point (30,10) is the solution. However, we should check the point by substituting it into the equations. Let's start with the first one. b=?-t+40t=30, b=1010=?-30+40Add terms10=10 Now, we'll check the second one. b=?-34t+50t=30, b=1010=?-34(30)+50Add terms12=12 Since both equations hold true, point (30,10) is a solution. This means that you should exercise 30 minutes on elliptical trainer and 10 minutes on stationary bike. | |

Exercises 28 We are told that you sold total of 28 candles. If we let s be the number of small candles sold and ℓ be the number of large candles sold, we can write an equation to represent it. 28=s+ℓ We can rearrange this equation, so it will be easier to plot it later. ℓ=-s+28 We also know that a large candle costs $6 and a small candle costs $4. With that information we can calculate how much express how much did you collect in terms of s and ℓ .Verbal expressionAlgebraic expression Price per large candle6 Number of large candles soldℓ Money collected from selling large candles6ℓ Price per small candle4 Number of small candles solds Money collected from selling small candles4s Total amount of money collected6ℓ+4sLet's build an equation that compare our expression with 144 given in the exercise. 6ℓ+4s=144 Isolating ℓ will help us graph this equation. 6ℓ+4s=144LHS−4s=RHS−4s6ℓ=-4s+144LHS/6=RHS/6ℓ=-32s+24 Now that we have both equations in the slope-intercept form we can plot them and find the point of interception.From the graph we can see that point (12,16) is the solution. However, we should check the point by substituting it into the equations. Let's start with the first one. ℓ=?-s+28s=12, ℓ=1616=?-12+28Add terms16=16 Now, we'll check the second one. ℓ=?-32s+24s=12, ℓ=1616=?-32(12)+24Multiply16=?-8+24Add terms16=16 Since both equations hold true, point (12,16) is a solution. This means that you sold 12 small candles and 16 large candles. | |

Exercises 29 We can calculate the area of a rectangle by multiplying the width w by the length ℓ. In the given diagram, we can see that w = 3x−3 and ℓ = 6. A=ℓ⋅ww=3x−3, ℓ=6A=6(3x−3)Distribute 6A=18x−18 The perimeter is calculated using the formula P=2ℓ+2w. Again, by substituting the known values for width and length, we can find a formula for the perimeter. P=2ℓ+2ww=3x−3, ℓ=6P=2(6)+2(3x−3) Simplify right-hand side Distribute 2P=2(6)+2(3x)−2(3)MultiplyP=12+6x−6Subtract terms P=6x+6 Finally, we can graph these equations and find their point of intersection.The solution is (2,18). This means that the perimeter and area are both 18 when x=2. | |

Exercises 30 Let's graph the given equation first. Next, we can find our friend's account balance after 6 months using the equation. Then we can write a linear equation that represents our account balance. Finally, we can interpret the slope and the y-intercept.Graphing the equation Luckily, the given equation is in already slope-intercept form. This means that we can identify the slope as 25 and the y-intercept as 250 right off the bat. Using this information, we can graph the equation.Balance account after 6 months To find our friends account balance after 6 months, let's substitute x=6 in the equation and simplify. y=25x+250x=6y=25⋅6+250 Simplify RHS Multiplyy=150+250Add terms y=400 After 6 months, our friend's account balance is $400.Our equation After 6 months, we want our account to have the same balance as our friend. This means it should pass through the point (6,400). Since we don't have more information about our bank account, let's assume that its y-intercept is 100. So far we have the function: y=ax+100. To find the slope, let's substitute (6,400) and isolate a. y=ax+100x=6, y=400400=a⋅6+100 Simplify RHS Multiply400=6a+100LHS−100=RHS−100300=6aLHS/6=RHS/650=aRearrange equation a=50 The slope is 50, so our equation becomes: y=50x+100.Interpreting slope and y-intercept Since x is the number of months and y is the bank account balance in dollars, the y-intercept tells us that at the beginning our account balance was 100. The slope is 50, which means that each increase by 1 to the right, the balance account increases by $50. Hence, we put $50 in our account every month. | |

Exercises 31 | |

Exercises 32 | |

Exercises 33 | |

Exercises 34 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. 10x+5y=5x+20LHS−10x=RHS−10x5y=-5x+20 LHS/5=RHS/5 LHS/5=RHS/5y=5-5x+20Write as a sum of fractionsy=5-5x+520Calculate quotienty=5-5x+4Cancel out common factors y=-x+4 | |

Exercises 35 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. 9x+18=6y−3xLHS+3x=RHS+3x12x+18=6y LHS/6=RHS/6 LHS/6=RHS/6612x+18=yWrite as a sum of fractions612x+618=yCalculate quotient 2x+3=yRearrange equationy=2x+3 | |

Exercises 36 To solve the given literal equation for y, we will use the Properties of Equality to apply inverse operations on the equation. Remember, only like terms can be combined. 43x+41y=5LHS−43x=RHS−43x41y=-43x+5 LHS⋅4=RHS⋅4 LHS⋅4=RHS⋅4y=4(-43x+5)Distribute 4 y=-3x+20 |

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

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