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Exercises 1 By graphing the equations, we can determine the point of intersection of the lines. The equations are both written in slope-intercept form, so we can immediately identify the slope m and y-intercept b. Equation (I): Equation (II): y=-3x+1 → m=-3, b=1y=x−7-3 → m=-1, b=-7 Let's graph each of the equations by plotting two points and drawing a line that passes through them. We will use the y-intercepts as the first points and determine the second points by using the slopes.The solution of this system of equations is the point of intersection of the two lines.We have shown that the solution is (2,-5). | |

Exercises 2 In this exercise, we are asked to graph the system and determine the number of solutions. We will tackle these tasks in that order.Graphing the System We can find the solution to a system of equations by graphing the lines. The point where the lines intersect is the solution to the system. The system is given as the following. {y=-4x+34x−2y=6(I)(II) Notice that the first line is written in the slope-intercept form, y=mx+b, where m is the slope of the line and b is the y-intercept. Let's rewrite the second line in this form. {y=-4x+34x−2y=6(I)(II)(II): LHS−4x=RHS−4x{y=-4x+3-2y=-4x+6(II): LHS/(-2)=RHS/(-2){y=-4x+3y=2x−3 Now, by analyzing the equations, we can identify the slopes m and y-intercepts b. Equation (I):Equation (II): m=-4andb=-3 m=-2andb=-3 We will graph the lines by plotting the y-intercept and then using the corresponding slope to find another point.Now that the lines are graphed, we can use the graph to determine the number of solutions.Number of solutions Notice from the graph above that the lines intersect once at the point (1,-1). This is the solution to the system. | |

Exercises 3 By graphing the given equations, we can determine the solution to the system. The solution will be the point at which the lines intersect. 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 5x+5y=15y=-1x+3-13 2x−2y=10y=1x+(-5)1-5 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.The lines intersect at the point (4,-1). This is the solution to the system. | |

Exercises 4 When solving a system of equations using substitution, there are three steps.Isolate a variable in one of the equations. Substitute the expression for that variable into the other equation and solve. Substitute this solution into one of the equations and solve for the value of the other variable. For this exercise, y is already isolated in one equation, so we can skip straight to solving! {3x+y=-9y=5x+7(I)(II)(I): y=5x+7{3x+5x+7=-9y=5x+7(I): Add terms{8x+7=-9y=5x+7(I): LHS−7=RHS−7{8x=-16y=5x+7(I): LHS/8=RHS/8{x=-2y=5x+7 Now, to find the value of y, we need to substitute x=-2 into either one of the equations in the given system. Let's use the second equation. {x=-2y=5x+7(II): x=-2{x=-2y=5(-2)+7(II): a(-b)=-a⋅b{x=-2y=-10+7(II): Add terms{x=-2y=-3 The solution, or point of intersection, to this system of equations is the point (-2,-3).Checking Our Answer To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct. {3x+y=-9y=5x+7(I)(II)(I), (II): x=-2, y=-3{3(-2)+(-3)=?-9-3=?5(-2)+7(I), (II): a(-b)=-a⋅b{-6−3=?-9-3=?-10+7(I), (II): Add and subtract terms{-9=-9-3=-3 Because both equations are true statements, we know that our solution is correct. | |

Exercises 5 In order to use the Substitution Method to solve a system of equations, we need to begin by isolating a variable in either of the equations. We can then substitute the expression for that variable into the second equation. We have been given the following system of equations. {x+4y=6x−y=1(I)(II) We will start by isolating the x variable in Equation (II). x−y=1⇔x=1+y Now we can substitute (II) into (I) and solve for y. {x+4y=6x=1+y(I)(II)(I): x=1+y{1+y+4y=6x=1+y(I): Add terms{1+5y=6x=1+y(I): LHS−1=RHS−1{5y=5x=1+y(I): LHS/5=RHS/5{y=1x=1+y We can now substitute y=1 into Equation (II) to find x. {y=1x=1+y(II): y=1{y=1x=1+1(II): Add terms{y=1x=2 The solution to the system is (2,1).Checking the solution In order to check the solution, we need to substitute 2 for x and 1 for y in both equations. If our solution is correct, we will end up with an identity. {x+4y=6x−y=1(I)(II)(I), (II): x=2, y=1{2+4⋅1=?62−1=?1(I): a⋅1=a{2+4=?62−1=?1(I), (II): Add and subtract terms{6=61=1 Our solution resulted in an identity in both equations, so it is correct. | |

Exercises 6 Using the Substitution Method to solve a system of equations requires us to begin by isolating either variable in one of the equations. We will then substitute the expression for that variable into the second equation. We have been given the following system of equations. {2x+3y=4y+3x=6(I)(II) Before we can substitute one equation into the other, we must alter one of the equations so that it is formatted as x=… or y=… Isolating the y term in (II) will require the fewest steps because its coefficient is 1. y+3x=6⇔y=-3x+6 Now we can substitute (II) into (I) and solve for x. {2x+3y=4y=-3x+6(I)(II)(I): y=-3x+6{2x+3(-3x+6)=4y=-3x+6(I): Distribute 3{2x−9x+18=4y=-3x+6(I): Subtract terms{-7x+18=4y=-3x+6(I): LHS−18=RHS−18{-7x=-14y=-3x+6(I): LHS/(-7)=RHS/(-7){x=2y=-3x+6 We can now substitute x=2 into either (I) or (II) to solve for y. We'll choose (II). {x=2y=-3x+6(II): x=2{x=2y=-3⋅2+6(II): (-a)b=-ab{x=2y=-6+6(II): Add terms{x=2y=0 The solution to the system of equations is (2,0).Checking the Solution Let's check this solution by substituting it back into both of the given equations. {2x+3y=4y=-3x+6(I)(II)(I): (II): x=2, y=0{2⋅2+3⋅0=?40=?-3⋅2+6(I)(II)Multiply{4+0=?40=?-6+6(I)(II)Add terms{4=40=0(I)(II) Both equations resulted in identities, so our solution to the system is correct. | |

Exercises 7 We are given total expenses of $20. We are also told that tubes of paint costs $4 and paintbrushes $0.5. Let's start by writing the first equation using word expressions. price per tube of paint⋅number of tubes of paint+price per paintbrush⋅number of paintbrushes=total cost If we let t represent the number of tubes of paint and p represent the number of paintbrushes we can write this equation algebraically. $4⋅t+$0.5⋅p=$20 Having that done, we can move on to the second equation. Since you buy twice as many brushes as tubes of paint, we have to multiply the number of tubes of paint by 2 for both sides to be equal. 2⋅number of tubes of paint=number of paintbrushes We have to use the same variables, t and p as in the first equation. 2⋅t=p Now we can combine both equations into a system of equations. {20=4t+0.5p2t=p(I)(II) Since variable p in the second equation is already isolated, it might be a good idea to solve the system by substitution {20=4t+0.5p2t=p(I)(II)(I): p=2t{20=4t+0.5(2t)2t=p(I)(II)(I): Multiply{20=4t+t2t=p(I)(II)(I): Add terms{20=5t2t=p(I)(II)(I): LHS/5=RHS/5{t=42t=p(I)(II) We have calculated that t=4, let's substitute it into the second equation and find the value of p. {t=42t=p(I)(II)(II): t=4{t=42(4)=p(I)(II)(II): Multiply{t=4p=8(I)(II) The solution to the system is t=4 and p=8. In the context of the word problem it means that you buy 4 tubes of paint and 8 paintbrushes. | |

Exercises 8 To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. {9x−2y=345x+2y=-6(I)(II) We can see that the y-terms will eliminate each other if we add (I) to (II). {9x−2y=345x+2y=-6(II): Add (I){9x−2y=345x+2y+(9x−2y)=-6+(34) (II): Solve for x (II): Add and subtract terms{9x−2y=3414x=28(II): LHS/14=RHS/14 {9x−2y=34x=2 Now we can now solve for y by substituting the value of x into either equation and simplifying. Let's use the first equation. {9x−2y=34x=2(I): x=2{9(2)−2y=34x=2 (I): Solve for y (I): Multiply{18−2y=34x=2(I): LHS−18=RHS−18{-2y=16x=2(I): LHS/-2=RHS/-2 {y=-8x=2 The solution, or intersection point, of the system of equations is (2,-8).Checking Our Answer To check this solution, we will substitute it back into the given system and simplify. If doing so results in true statements for every equation in the system, our solution is correct. {9x−2y=345x+2y=-6(I)(II)(I), (II): x=2, y=-8{9(2)−2(-8)=?345(2)+2(-8)=?-6(I), (II): Multiply{18+16=?3410−16=?-6(I), (II): Add and subtract terms{34=34-6=-6 Because both equations are true statements, we know that our solution is correct. | |

Exercises 9 We will find the solution of the system of equations and then we will check the answer.Finding the solution To use the Elimination Method, either the x- or y-terms in both equations must cancel out when the equations are added. If we multiply the second equation by 2, the coefficients of y will be 6 and -6. Then, when we add the equations, y will cancel out, and we can solve for x. {x+6y=282x−3y=-19(I)(II)(II): LHS⋅2=RHS⋅2{x+6y=284x−6y=-38(II): Add (I){x+6y=284x−6y+x+6y=-38+28(II): Add terms{x+6y=285x=-10(II): LHS/5=RHS/5{x+6y=28x=-2 Now, we can substitute x=-2 into Equation (I) and solve for y. {x+6y=28x=-2(I): x=-2{-2+6y=28x=-2(I): LHS+2=RHS+2{6y=30x=-2(I): LHS/6=RHS/6{y=5x=-2 The solution to the system of equations, which is the point of intersection of the two lines, is (-2,5).Checking the answer To check the answer, we will substitute -2 for x and 5 for y in both equations and simplify. {x+6y=282x−3y=-19(I)(II)(I), (II): x=-2, y=5{-2+6⋅5=?282(-2)−3⋅5=?-19 Simplify LHS (II): a(-b)=-a⋅b{-2+6⋅5=?28-4−3⋅5=?-19(I), (II): Multiply{-2+30=?28-4−15=?-19(I), (II): Add and subtract terms {28=28-19=-19 Since we have arrived to two identities, we know that the solution (-2,5) is correct. | |

Exercises 10 To use the Elimination Method, either the x- or y-terms in both equations must cancel out when the equations are added. If we multiply the first equation by -5 and the second equation by 7, the coefficients of y will be 35 and -35. {8x−7y=-36x−5y=-1(I)(II)(I): LHS⋅(-5)=RHS⋅(-5){-40x+35y=156x−5y=-1(II): Multiply by 7{-40x+35y=1542x−35y=-7 Now when we add the equations, the y-terms will cancel out and we can solve for x. {-40x+35y=1542x−35y=-7(II): Add (I){-40x+35y=1542x−35y+(-40x)+35y=-7+15(II): Add and subtract terms{-40x+35y=152x=8(II): LHS/2=RHS/2{-40x+35y=15x=4 Substitute x=4 into Equation (I) and solve for y. {-40x+35y=15x=4(I): x=4{-40⋅4+35y=15x=4(I): Multiply{-160+35y=15x=4(I): LHS+160=RHS+160{35y=175x=4(I): LHS/35=RHS/35{y=5x=4 The solution to the system of equations is (4,5). Let's check this solution by substituting it back into both of the given equations. {8x−7y=-36x−5y=-1(I)(II)(I): (II): x=4, y=5{8⋅4−7⋅5=?-36⋅4−5⋅5=?-1(I)(II)Multiply{32−35=?-324−25=?-1(I)(II)Subtract terms{-3=-3-1=-1(I)(II) Both equations resulted in identities, so our solution to the system is correct. | |

Exercises 11 Because the variable x is already isolated in the first equation, it is easiest to solve by the Substitution Method. Let's do it! {x=y+2-3x+3y=6(I)(II)(II): x=y+2{x=y+2-3(y+2)+3y=6(II): Distribute -3{x=y+2-3y−6+3y=6(II): Add terms{x=y+2-6=6 We have reached a contradiction, -6 can never be equal to 6. This means that there is no solution to the system of equations. Why might this be? Let's isolate y in Equation (I) and Equation (II) in order to compare them. {x=y+2-3x+3y=6(I)(II) Isolate y in both equations (I): LHS−2=RHS−2{x−2=y-3x+3y=6(I): Rearrange equation{y=x−2-3x+3y=6(II): LHS+3x=RHS+3x{y=x−23y=6+3x(II): LHS/3=RHS/3{y=x−2y=2+x(II): Rearrange equation {y=x−2y=x+2 When we look at both equations in the slope-intercept form, we can see that they have the same slope and different y-intercepts. This means that they are parallel lines and will never intersect. That is why there are no solutions to the system of equations. | |

Exercises 12 In order to solve the system of equations, we should begin by rewriting the system of equations in slope-intercept form. {3x−6y=-9-5x+10y=10(I)(II)(I): LHS−3x=RHS−3x{-6y=-3x−9-5x+10y=10(I): LHS/(-6)=RHS/(-6){y=21x+23-5x+10y=10(II): LHS+5x=RHS+5x{y=21x+2310y=5x+10(II): LHS/10=RHS/10{y=21x+23y=21x+1 Both lines have a slope of 21. Therefore, we can say that these lines are parallel. Let's graph them to see clearly.Parallel lines written as a system of equations will always result in an answer of no solution. This is because parallel lines, by definition, will never intersect. | |

Exercises 13 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the easiest approach. Instead, we'll use the elimination method. To use the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {-4x+4y=323x+24=3y(I)(II) ⇒{-4x+4y=323x−3y=-24(I)(II) In its current state, this won't happen. Therefore, we need to find a common multiple between two variable like-terms in the system. If we multiply (I) by 3 and multiply (II) by 4, both the x-terms and the y-terms will have opposite coefficients. {3(-4x+4y)=3(32)4(3x−3y)=4(-24) ⇒ {-12x+12y=9612x−12y=-96 We can see that both the x-terms and y-terms will eliminate each other if we add (I) to (II). {-12x+12y=9612x−12y=-96(II): Add (I){-12x+12y=9612x−12y+(-12x+12y)=-96+(96)(II): Add and subtract terms{-12x+12y=960=0 Solving this system of equations resulted in an identity; 0 is always equal to itself. The lines are the same and have infinitely many intersection points. Therefore, the system has infinitely many solutions. | |

Exercises 14 To graph the equation 31x+5=-2x−2, we create will two functions out of the left and right hand sides of the equation. y=31x+5and y=-2x−2 The x-coordinate where the graphs of these functions intersect is the solution to our equation. Fortunately, our functions are already in a slope-intercept form, so we can move straight to graphing using the y-intercepts and the slopes of the lines.The graphs intersect at x=-3, which is our solution. We can check our solution by substituting x=-3 into the original equation. 31x+5=-2x−2x=-331(-3)+5=?-2(-3)−2Multiply-1+5=?6−2Add and subtract terms4=4 The resulting statement is true, so our solution is correct. | |

Exercises 15 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−9∣∣x+1∣=-(-x−9∣x+1∣=-(-x−9) 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+1and y=-x−9 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,-4) which means the equation's solution is x=-5. Let's check whether it's correct by substituting it into the original equation. ∣x+1∣=∣-x−9∣x=-5∣-5+1∣=?∣-(-5)−9∣ Simplify -(-a)=a∣-5+1∣=?∣5−9∣Add and subtract terms∣-4∣=?∣-4∣∣-4∣=4 4=4 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=x+1and y=-(-x+9) ⇒y=x+9 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. Therefore, the only solution to this equation is x=-5. | |

Exercises 16 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−8∣= ∣x+5∣∣2x−8∣=-(x+5∣2x−8∣=-(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 and right hand sides of the equation. y=2x−8and y=x+5 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 (13,18) which means the equation's solution is x=13. Let's check whether it's correct by substituting it into the original equation. ∣2x−8∣=∣x+5∣x=13∣2(13)−8∣=?∣13+5∣ Simplify Multiply∣26−8∣=?∣13+5∣Add and subtract terms∣18∣=?∣18∣∣18∣=18 18=18 The equation is true, so x=13 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−8and y=-(x+5) ⇒y=-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 (1,-6), which means the solution to this equation is x=1. ∣2x−8∣=∣x+5∣x=1∣2(1)−8∣=?∣1+5∣ Simplify Multiply∣2−8∣=?∣1+5∣Add and subtract terms∣-6∣=?∣6∣∣-6∣=66=?∣6∣∣6∣=6 6=6 Checking our solution, we confirmed that x=1 is also a solution. | |

Exercises 17 To graph the inequality, we must first graph its boundary line and then shade the region that contains the solution set. For the inequality y>-4 the boundary line is given by the equation y=-4, which is a horizontal line passing through (0,-4) on the y-axis. Be aware that we will use a dashed line since we have a strict inequality.The inequality y>-4 describes all possible values for y that are greater than -4. To show this on the graph we need to shade the area above the line, since all these points have y-coordinates greater than -4. | |

Exercises 18 Graphing an inequality involves three steps.Plotting the boundary line. Testing a point to determine which side of the graph is the solution set. Shading one side of the graph to show the solution set.Boundary Line To graph the inequality, we first have to draw the boundary line. To do this, we treat the inequality as if it was an equation. -9x+3y≥3⇒-9x+3y=3 Before we can graph it, we need to rewrite the equation in slope-intercept form. -9x+3y=3LHS+9x=RHS+9x3y=9x+3LHS/3=RHS/3y=3x+1 Now that the equation is in slope-intercept form, we can identify the slope and y-intercept. Slope:my-intercept: b=-3=-1 We will use the y-intercept and the slope to graph two points and connect them with a line. In this case, because the inequality is non-strict, the boundary line should be solid.Testing a Point Now we have to determine which side of the boundary line we should shade. Let's substitute a test point that is not on the boundary line into the inequality. If the substitution creates a true statement, we shade the region that includes the test point. If not, we shade the opposite region. Let's use (0,0). y≥3x+1x=0, y=00≥?3(0)+1Zero Property of Multiplication0≥?0+1Add terms0≱1 Since the substitution of the test point did not create a true statement, we will shade the region that does not contain the point. | |

Exercises 19 Graphing an inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.Boundary Line To graph the inequality, we have to draw the boundary line. The equation of a boundary line is written by replacing the inequality symbol from the inequality with an equals sign. Inequality5x+10y < 40 Boundary Line 5x+10y = 40 To draw this line, we will first rewrite the equation in slope-intercept form. 5x+10y=40LHS−5x=RHS−5x10y=-5x+40LHS/10=RHS/10y=-21x+4 Now that the equation is in slope-intercept form, we can identify the slope m and y-intercept (0,b). y=-21x+4 We will plot the y-intercept (0,4), then use the slope m=-21 to plot another point on the line. Connecting these points with a dashed line will give us the boundary line of our inequality. Note that the boundary line is dashed, not solid, because the inequality is strict.Shading the Plane To decide which side of the boundary line to shade, we will substitute a test point that is not on the boundary line into the given inequality. If the substitution creates a true statement, we shade the region that includes the test point. Otherwise, we shade the opposite region. Let's use (0,0) as our test point. 5x+10y<40x=0, y=05(0)+10(0)<?40Zero Property of Multiplication0+0<?40Add terms0<40 Since the substitution of the test point created a true statement, we will shade the region that contains the point. | |

Exercises 20 We have been given the following system of inequalities and asked to graph its solution set. {y≤x−3y≥x+1(I)(II) In order to do so, we first need to identify the boundary lines for Inequality (I) and Inequality (II).Inequality (I) In this case, this boundary line will be y=x−3, where the slope is 1 and the y-intercept is -3. Note that this boundary line will be solid because the inequality is less than or equal to.To choose which side of the boundary line should be shaded, we will test the point (0,0). y≤x−3x=0, y=00≤?0−3Subtract term0≰-3 Because 0≰-3, we will shade the region which does not contain the point (0,0).Inequality (II) Our second boundary line will be y=x+1, where the slope is 1 and the y-intercept is also 1. Note that this boundary line will also be solid because the inequality is greater than or equal to. To decide which region should be shaded, we test the point (0,0). y≥x+1x=0, y=00≥?01Add terms0≱1 Because 0≱1, we will shade the region which does not contain (0,0).Final solution As you can see in the last graph, the half-planes do not intersect each other. Therefore, the system of inequalities has no solution. | |

Exercises 21 The solution to a system of inequalities can be found graphically. To graph the inequalities we must draw the boundary line and shade the region on the coordinate plane that shows the solution. Since our inequalities are already in slope-intercept form, {y>-2x+3y≥41x−1(I)(II) we will graph each inequality separately. Let's start with Inequality (I).Inequality (I) To graph an inequality, we must draw the boundary line and then shade the appropriate region. The boundary line for Inequality (I) is y=-2x+3. Since this is written in slope-intercept form, we know that the slope of this line is -2 and the y-intercept is (0,3). Note that the inequality symbol is >, so the line will be dashed.To determine which side of the inequality to shade, we will substitute a test point into the inequality. If the point makes a true statement, the point is a solution and we should shade the region containing the point. Otherwise, we shade the opposite region. We'll test (0,0). y>-2x+3x=0, y=00>?-2⋅0+3Use the Zero Product Property0>?0+3Add terms0≯3 Since 0 is not greater than 3, (0,0) is not a solution to the inequality. Thus, we'll shade the region to the right of the boundary line.Let's move on to Inequality (II).Graphing Inequality (II) We will follow the same process to graph Inequality (II). The equation of boundary line is y=41x−1. The slope is 41 and the y-intercept is -1. The boundary line will be solid because the inequality symbol is ≥. Lastly, we can test (0,0) to determine which region to shade. y≥41x−1x=0, y=00≥?41⋅0−1Use the Zero Product Property0≥?0−1Subtract term0≥-1 Since 0 is greater than -1, (0,0) is a solution to the inequality. Thus, we'll shade the region to the left of the boundary line. We can now add the graph of Inequality (II) to the graph of Inequality (I).The solution to this system of inequalities is the area where the shading overlaps. We distinguish this area in the graph below. | |

Exercises 22 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {x+3y>62x+y<7(I)(II) The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether or not the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slope m and y-intercept b of each boundary line.Let's find each of these key pieces of information for the inequalities in the system.InformationInequality (I)Inequality (II) Given Inequalityx+3y>62x+y<7 Boundary Line Equationx+3y=62x+y=7 Solid or Dashed?> ⇒ Dashed< ⇒ Dashed y=mx+by=-31x+2y=-2x+7 With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.Let's substitute the point (0,0) for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityx+3y>62x+y<7 Substitute (0,0)0+3(0)>?62(0)+(0)<?7 Simplify0≯60<7 Shaded Regionoppositesame For Inequality (I) we will shade the region opposite our test point, or above the boundary line. For Inequality (II), we will shade the region containing the test point, or below the boundary line. |

##### 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
- Solving Equations by Graphing
- Graphing Linear Inequalities in Two Variables
- Systems of Linear Inequalities
- Chapter Test
- Cumulative Assessment