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Exercises 1 To solve a system of linear equations by substitution, follow the three steps below and all will be fine:Isolate one of the variables in one of the equations. Substitute the corresponding expression for this variable into the other equation and solve. Having solved for one of the variables, substitute this value back into the first equation to find the remaining variable. | |

Exercises 2 We can decide which variable to solve for by looking at the coefficients. There are a few cases to consider.If either equation has a variable with a coefficient of 1, isolating this will likely be the easiest. This can be done without having to multiply or divide. If there is no variable with a coefficient 1, but there is a coefficient of -1, we should choose to solve for this. We just have to change the signs of the entire equation before we isolate the variable. If no equations have a coefficient of 1 or -1, isolate the variable that will give you the least amount of grief. For example, in the equation7y−63x=28 it is better to isolate y as this will leave us with an equation without any fractions.y=9x+4 vs. x=9y−94 | |

Exercises 3 To a system of equations by substitution, a variable must be isolated on one side of one of the equations. Looking at the given equations, we see that it will be necessary to apply the Properties of Equality for this to happen. {x+4y=30x−2y=0(I)(II) Which variable in which equation would be the easiest to isolate? Well, the coefficient of x in both equations is 1, which makes it easy to isolate x. {x=30−4yx=2y(I)(II) We should also consider how difficult the substitution will become when we solve the system. The expression equal to x in (I) is more complicated than the expression equal to x in (II), so we should use (II) to solve for a variable. | |

Exercises 4 To solve a system of equations by substitution, a variable must be isolated on one side of one of the equations. Looking at the given equations, we see that it will be necessary to apply the Properties of Equality for this to happen. {3x−y=02x+y=−10(I)(II) Which variable in which equation would be the easiest to isolate? Well, the coefficient of y in the equations is 1 or -1, which makes it easy to isolate y. {3x=yy=−2x−10 If we rearrange the first equation, we get: {y=3xy=−2x−10 We should also consider how difficult the substitution will become when we solve the system. The expression equal to y in (II) is more complicated than the expression equal to y in (I), so we should use (I) to solve for a variable. | |

Exercises 5 We have to discuss what to isolate and in which equation.What should we isolate? Often, you might think that we must completely isolate a variable. HOWEVER, to perform a substitution, it's enough that the equations have some matching part. If we look at the system of equations, we can see that they both contain 5x: {5x+3y=115x−y=5 If we isolate this term in one of the equations, we can substitute for 5x and solve for y. Therefore, we should choose to isolate 5x.Which equation should we choose? Comparing the equations, we can see that in the second equation y has a coefficient of 1 and there is a constant 5. Conversely, in the first equation, the coefficient of y is 3 and the constant is 11. The fact that we have easier, smaller numbers in the second equation makes this the best candidate. | |

Exercises 6 Comparing the equations, we can see that the second equation has coefficients of 1 for both the x and y. Therefore, this equation is easier to isolate one of the variables. Since both of the variables have a coefficient of 1, we can choose to isolate either. | |

Exercises 7 To solve a system of equations by substitution, a variable must be isolated on one side of one of the equations. Looking at the given equations, we see that it will be necessary to apply the Properties of Equality for this to happen. {x−y=-34x+3y=-5(I)(II) Which variable in which equation would be the easiest to isolate? Well, the coefficient of x in the first equation is 1, which makes it easy to isolate x. {x=y−34x+3y=-5 Note, that it is impossible to isolate the variable in the equation (II) in one operation, because the coefficient of x in (II) is 4. For that reason we should use (I) to solve for a variable. | |

Exercises 8 To solve a system of equations by substitution, a variable must be isolated on one side of one of the equations. Looking at the given equations, we see that it will be necessary to apply the Properties of Equality for this to happen. {3x+5y=25x−2y=-6(I)(II) Which variable in which equation would be the easiest to isolate? Well, the coefficient of x in the second equation is 1, which makes it easy to isolate x. {3x+5y=25x=2y−6 Note, that it is impossible to isolate the variable in the equation (I) in one operation, because the coefficient of x in (I) is 3. For that reason we should use (I) to solve for a variable. | |

Exercises 9 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, x is already isolated in one equation, so we can skip straight to solving! {x=17−4yy=x−2(I)(II)(II): y=x−2{x=17−4yy=17−4y−2(II): Subtract term{x=17−4yy=15−4y(II): LHS+4y=RHS+4y{x=17−4y5y=15(II): LHS/5=RHS/5{x=17−4yy=3 Great! Now, to find the value of x, we need to substitute y=3 into either one of the equations in the given system. Let's use the first equation. {x=17−4yy=3(I): y=3{x=17−4(3)y=3(I): Multiply{x=17−12y=3(I): Subtract term{x=5y=3 The solution, or point of intersection, to this system of equations is the point (5,3).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. {x=17−4yy=x−2(I), (II): x=5, y=3{5=?17−4(3)3=?5−2(II): Multiply{5=?17−123=?5−2(I), (II): Subtract term{5=53=3 Because both equations are true statements, we know that our solution is correct. | |

Exercises 10 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! {x=17−4yy=x−2(I)(II)(I): y=x−2{x=17−4(x−2)y=x−2(I): Distribute 4{x=17−4x+8y=x−2(I): Add terms{x=25−4xy=x−2(I): LHS+4x=RHS+4x{5x=25y=x−2(I): LHS/5=RHS/5{x=5y=x−2 Great! Now, to find the value of y, we need to substitute x=5 into either one of the equations in the given system. Let's use the second equation. {x=5y=x−2(II): x=5{x=5y=5−2(II): Subtract term{x=5y=3 The solution, or point of intersection, to this system of equations is the point (5,3).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. {x=?17−4yy=?x−2(I), (II): x=5, y=3{5=?17−4(3)3=?5−2(I): Multiply{5=?17−123=?5−2(I), (II): Subtract terms{5=53=3 Because both equations are true statements, we know that our solution is correct. | |

Exercises 11 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! {6x−9=yy=-3x(I)(II)(I): y=-3x{6x−9=-3xy=-3x(I): LHS−6x=RHS−6x{-9=-9xy=-3x(I): Rearrange equation{-9x=-9y=-3x(I): LHS/-9=RHS/-9{x=1y=-3x Great! Now, to find the value of y, we need to substitute x=1 into either one of the equations in the given system. Let's use the second equation. {x=1y=-3x(II): x=1{x=1y=-3(1)(II): Multiply{x=1y=-3 The solution, or point of intersection, to this system of equations is the point (1,-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. {6x−9=?yy=?-3x(I), (II): x=1, y=-3{6(1)−9=?-3-3=?-3(1)(I), (II): Multiply{6−9=?-3-3=?-3(I): Subtract term{-3=-3-3=-3 Because both equations are true statements, we know that our solution is correct. | |

Exercises 12 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, x is already isolated in one equation, so we can skip straight to solving! {x=16−4y3x+4y=8(I)(II)(II): x=16−4y{x=16−4y3(16−4y)+4y=8(II): Distribute 3{x=16−4y48−12y+4y=8(II): Add terms{x=16−4y48−8y=8(II): LHS−48=RHS−48{x=16−4y-8y=-40(II): LHS/-8=RHS/-8{x=16−4yy=5 Great! Now, to find the value of x, we need to substitute y=5 into either one of the equations in the given system. Let's use the first equation. {x=16−4yy=5(I): y=5{x=16−4(5)y=5(I): Multiply{x=16−20y=5(I): Subtract term{x=-4y=5 The solution, or point of intersection, to this system of equations is the point (-4,5).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. {x=?16−4y3x+4y=?8(I)(II)(I), (II): x=-4, y=5{-4=?16−4(5)3(-4)+4(5)=?8(I), (II): Multiply{-4=?16−20-12+20=?8(I), (II): Add and subtract terms{-4=-48=8 Because both equations are true statements, we know that our solution is correct. | |

Exercises 13 Analyzing the system of equations, we can see that it's easier to isolate x by dividing by 2 in the first equation. Then we can substitute the value of x into the second equation and solve for y. {2x=12x−5y=-29(I)(II)(I): LHS/2=RHS/2{x=6x−5y=-29(II): x=6{x=66−5y=-29 (II): Solve for y (II): LHS−6=RHS−6{x=6-5y=-35(II): Change signs{x=65y=35LHS/5=RHS/5 {x=6y=7 The solution is (6,7). Let's check our solution by substituting it into each equation. {2x=12x−5y=-29(I)(II)(I): x=6, y=7{2⋅6=?126−5⋅7=?-29(I), (II): Multiply{12=?126−35=?-29(II): Subtract term{12=12-29=-29 Both equations are true so (6,7) is the solution to the system of equations. | |

Exercises 14 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. Let's start with isolating x in the equation II. {2x−y=23x−9=-1(I)(II)(II): LHS+9=RHS+9{2x−y=23x=8 Now that we've isolated x, we can solve the system by substitution. {2x−y=23x=8(II): x=8{2(8)−y=23x=8(II): Multiply{16−y=23x=8(II): LHS−16=RHS−16{-y=7x=8(II): LHS/-1=RHS/-1{y=-7x=8 The solution, or point of intersection, to this system of equations is the point (8,-7).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. {2x−y=23x−9=-1(I)(II)(I), (II): x=8, y=-7{2(8)−(-7)=?238−9=?-1(II): Multiply{16−(-7)=?238−9=?-1(I), (II): Subtract term{23=23-1=-1 Because both equations are true statements, we know that our solution is correct. | |

Exercises 15 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. Observing the given equations, it looks like it will be simplest to isolate x in the equation II. {5x+2y=9x+y=-3(I)(II)(II): LHS−y=RHS−y{5x+2y=9x=-3−y Now that we've isolated x, we can solve the system by substitution. {5x+2y=9x=-3−y(I): x=-3−y{5(-3−y)+2y=9x=-3−y(I): Multiply{-15−5y+2y=9x=-3−yx=-3−y(I): Simplify terms{-15−3y=9x=-3−y(I): LHS+15=RHS+15{-3y=24x=-3−y(I): LHS/-3=RHS/-3{y=-8x=-3−y Great! Now, to find the value of x, we need to substitute y=-8 into either one of the equations in the given system. Let's use the second equation. {y=-8x=-3−y(I): y=-8{y=-8x=-3−(-8)(I): Subtract terms{y=-8x=5 The solution, or point of intersection, to this system of equations is the point (5,-8).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. {5x+2y=9x+y=-3(I), (II): x=5, y=-8{5(5)+2(-8)=?95+(-8)=?-3(II): Multiply{25−16=?95+(-8)=?-3(I), (II): Add and subtract terms{9=9-3=-3 Because both equations are true statements, we know that our solution is correct. | |

Exercises 16 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. Observing the given equations, it looks like it will be simplest to isolate x in the equation II. {11x−7y=-14x−2y=-4(I)(II)(II): LHS+2y=RHS+2y{11x−7y=-14x=-4+2y Now that we've isolated x, we can solve the system by substitution. {11(-4+2y)−7y=-14x=-4+2y(I): Distribute 11{-44+22y−7y=-14x=-4+2y(I): Simplify terms{-44+15y=-14x=-4+2y(I): LHS+44=RHS+44{15y=30x=-4+2y(I): LHS/15=RHS/15{y=2x=-4+2y Great! Now, to find the value of x, we need to substitute y=2 into either one of the equations in the given system. Let's use the first equation. {y=2x=-4+2y(II): y=2{y=2x=-4+2(2)(II): Multiply{y=2x=-4+4(II): Add terms{y=2x=0 The solution, or point of intersection, to this system of equations is the point (0,2).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. {11x−7y=-14x−2y=-4(I), (II): x=0, y=2{11(0)−7(2)=?-140−2(2)=?-4(II): Multiply{-14=-14-4=-4 Because both equations are true statements, we know that our solution is correct. | |

Exercises 17 In the first step, no mistakes have been made. Moving on to the second step, the expression for y should be substituted into the other equation to calculate x. Not the same equation! Let's complete the second step properly. 8x+2y=-12y=5x−48x+2(5x−4)=-12 Solve for x Distribute 28x+10x−4=-12Add terms18x−8=-12LHS+8=RHS+818x=-4LHS/18=RHS/18x=18-4Put minus sign in front of fractionx=-184ba=b/2a/2 x=-92 The value of x is -92. | |

Exercises 18 In step 1, y was isolated correctly. In step 2, the expression from step 1 was substituted into the second equation and x was solved for correctly. In step 3, y was substituted with the value of x. This is the mistake. To solve it correctly, we should substitute the x with 6 and calculate the value of y. 3x+y=9x=63⋅6+y=9Multiply18+y=9LHS−18=RHS−18y=-9 The value of y is -9. | |

Exercises 19 Let c be the number of acres of corn and w be the number of acres of wheat the farmer wants to plant. Since there is a total of 180 acres, we can write the first equation as follows. c+w=180 We also know that the farmer wants to plant three times as many acres of corn as wheat. We can write this as c=3w. If we combine the equations, we have the following system of equations. {c+w=180c=3w To find the number of acres of each crop the farmer should plant, we have to solve for c and w. Since c is already isolated in the second equation, we will substitute 3w for c in the first equation and proceed to solve for w. {c+w=180c=3w(I)(II)(I): c=3w{3w+w=180c=3w(I): Add terms{4w=180c=3w(I): LHS/4=RHS/4{w=45c=3w Now, let's substitute w for 45 in Equation (II) to find c. {w=45c=3w(II): w=45{w=45c=3⋅45(II): Multiply{w=45c=135 Therefore, the farmer should plant 45 acres of wheat and 135 acres of corn. | |

Exercises 20 Let's start by defining the variables. Let p be the number of personal tubes and c the number of cooler tubes. Since we are told that the group rents a total of 15 tubes, the sum of p and c must be 15. p+c=15 We also know that the group spends $270 in renting the tubes. Additionally, the cost of a one-person tube is $20, and the cost of renting a cooler tube is $12.50. We can write our second equation using this information. 20p+12.5c=270 The equations combined together for a system of linear equations. {p+c=1520p+12.5c=270(I)(II) To find how many of each type of tube the group rents, we have to solve for p and c. To do so, we will use the Substitution Method. Let's start by isolating the p-variable in Equation (I). {p+c=1520p+12.5c=270(I): LHS−c=RHS−c{p=15−c20p+12.5c=270 Now we can substitute 15−c for p in Equation (II). {p=15−c20p+12.5c=270(II): p=15−c{p=15−c20(15−c)+12.5c=270 Solve for c (II): Distribute 20{p=15−c300−20c+12.5c=270(II): Add terms{p=15−c300−7.5c=270(II): LHS−300=RHS−300{p=15−c-7.5c=-30(II): LHS/(-7.5)=RHS/(-7.5) {p=15−cc=4 Now we can find p by substituting 12 for c in Equation(I). {p=15−cc=4(I): c=4{p=15−4c=4(I): Subtract term{p=11c=4 The group rents 11 personal tubes and 4 cooler tubes. | |

Exercises 21 We are asked to find a system of equations with (3,5) as its solution. For our first equation, we can choose any arbitrary slope and then substitute it, together with x=3 and y=5, in the general formula for the slope-intercept form of a line. Let's choose a slope of 2. y−y1=m(x−x1)Substitute valuesy−5=2(x−3)Distribute 2y−5=2x−6LHS+5=RHS+5y=2x−1 We can find the second equation in the same way, as long as we do not choose the same slope! Let's now choose a slope of 5. Again, we know the line passes through (3,5), and so this becomes our point. y−y1=m(x−x1)Substitute valuesy−5=5(x−3)Distribute 5y−5=5x−15LHS+5=RHS+5y=5x−10 When we combine these equations, we form a system of equations which has (3,5) as its solution. {y=2x−1y=5x−10 | |

Exercises 22 We have been asked to find any system of equations with (-2,8) as its solution. For our first equation, we will choose an arbitrary slope of 2. Then we can substitute this slope and the given solution into the point-slope form. y−y1=m(x−x1)Substitute valuesy−8=2(x−(-2)) Solve for y Remove parentheses and change signsy−8=2(x+2)Multiplyy−8=2x+4LHS+8=RHS+8 y=2x+12 We can find the second equation in the same way. Let's now choose a slope of 5. Again, we know it has to pass through (-2,8), so this becomes our point. y−y1=m(x−x1)Substitute valuesy−8=5(x−(-2)) Solve for y Remove parentheses and change signsy−8=5(x+2)Multiplyy−8=5x+10LHS+8=RHS+8 y=5x+18 When we combine these equations, we form a system of equations which has (-2,8) as its solution. {y=2x+12y=5x+18 | |

Exercises 23 We have been asked to find any system of equations with (-4,-12) as its solution. For our first equation, we will choose an arbitrary slope of 2. Then, we can substitute this slope and the given solution into the point-slope form. y−y1=m(x−x1)Substitute valuesy−(-12)=2(x−(-4)) Solve for y Remove parentheses and change signsy+12=2(x+4)Distribute 2y+12=2⋅x+2⋅4Multiplyy+12=2x+8LHS−12=RHS−12 y=2x−4 We can find the second equation in the same way. Let's now choose a slope of 5. Again, we know it has to pass through (-4,-12), so this becomes our point. y−y1=m(x−x1)Substitute valuesy−(-12)=5(x−(-4)) Solve for y Remove parentheses and change signsy+12=5(x+4)Distribute 5y+12=5⋅x+5⋅4Multiplyy+12=5x+20LHS−12=RHS−12 y=5x+8 When we combine these equations, we form a system of equations which has (-4,-12) as its solution. {y=2x−4y=5x+8 | |

Exercises 24 We are asked to find a system of equations with (15,-25) as its solution. For our first equation, we can choose any arbitrary slope and then substitute it, together with x=15 and y=-25, in the general formula for the slope-intercept form of a line. Let's choose a slope of 2. y−y1=m(x−x1)Substitute valuesy−(-25)=2(x−15) Solve for y a−(-b)=a+by+25=2(x−15)Distribute 2y+25=2x−30LHS−25=RHS−25 y=2x−55 We can find the second equation in the same way, as long as we do not choose the same slope! Let's now choose a slope of 3. Again, we know the line passes through (15,-25), and so this becomes our point. y−y1=m(x−x1)Substitute valuesy−(-25)=3(x−15) Solve for y a−(-b)=a+by+25=3(x−15)Distribute 3y+25=3x−45LHS−25=RHS−25 y=3x−70 When we combine these equations, we form a system of equations which has (15,-25) as its solution. {y=2x−55y=3x−70 | |

Exercises 25 In order to find the solution to the given problem, we will write two equations and solve the system formed.Writing the System Let x be the number of five-point problems and y the number of two-point problems. Knowing that the test consists of 38 problems, the addition of x and y must be 38. x+y=38 Let's now write our second equation. The expression 5x represents the number of points obtained from five-point questions, and 2y represents the points obtained from two-point questions. The test is worth 100 points. Therefore, the addition of these two expressions must equal 100. 5x+2y=100 With our two equations, we can form a system. {x+y=385x+2y=100(I)(II)Solving the System Note that the y-variable in Equation (I) can be isolated in just one step. {x+y=385x+2y=100⇔{y=38−x5x+2y=100(I)(II) Since one variable is isolated, we will use the Substitution Method. {y=38−x5x+2y=100(II): y=38−x{y=38−x5x+2(38−x)=100 (II): Solve for x (II): Distribute 2{y=38−x5x+76−2x=100(II): Subtract term{y=38−x3x+76=100(II): LHS−76=RHS−76{y=38−x3x=24(II): LHS/3=RHS/3 {y=38−xx=8 Now that we know that x=8, we will substitute 8 for x in Equation (I), and solve for y. {y=38−xx=8(I): x=8{y=38−8x=8(I): Subtract term{y=30x=8 The solution to the system, which is the point of intersection of the line, is (8,30). In the context of the problem, this means that there are 8 five-point problems and 30 two-point problems in the test. | |

Exercises 26 To solve the given problem, we will form two equations and combine them into a system. | |

Exercises 27 aThe sum of a triangle's interior angles is 180∘. From the figure, we can see that one of the angles is a right angle. The remaining two are x and y. By adding these together, we can equate their sum with 180. x+y+90=180bUsing the equation found in Part A and the given equation, we can form a system of equations. {x+y+90=180x+2=3y(I)(II) Let's solve it using the Substitution Method. To do so, we will start by isolating the x-variable in Equation (I). {x+y+90=180x+2=3y(I)(II)(I): LHS−90=RHS−90{x+y=90x+2=3y(I): LHS−y=RHS−y{x=90−yx+2=3y Let's now substitute 90−y for x in Equation (II). {x=90−yx+2=3y(II): x=90−y{x=90−y90−y+2=3y (II): Solve for y (II): Add terms{x=90−y92−y=3y(II): LHS+y=RHS+y{x=90−y92=4y(II): LHS/4=RHS/4{x=90−y23=y(II): Rearrange equation {x=90−yy=23 Finally, to find the value of x, we will substitute 23 for y in Equation (I). {x=90−yy=23(I): y=23{x=90−23y=23(I): Subtract term{x=67y=23 The solution to the system of equations, which is the point of intersection of the lines, is (67,23). In the context of the problem, this means that the measures of the missing angles are 67∘ and 23∘. | |

Exercises 28 aThe sum of a triangle's interior angles is 180∘. From the figure, we can see that one of the angles is y−18. The remaining two are x and y. By adding these together, we can equate their sum with 180. x+y+y−18=180bUsing the equation found in Part A and the given equation, we can form a system of equations. {x+y+y−18=1803x−5y=-22(I)(II) Let's solve it using the Substitution Method. To do so, we will start by isolating the x-variable in Equation (I). {x+y+y−18=1803x−5y=-22 (I): Isolate x (I): Add terms{x+2y−18=1803x−5y=-22(I): LHS+18=RHS+18{x+2y=1983x−5y=-22(I): LHS−2y=RHS−2y {x=198−2y3x−5y=-22 Let's now substitute 198−2y for x in Equation (II). {x=198−2y3x−5y=-22(II): x=198−2y{x=198−2y3(198−2y)−5y=-22 (II): Solve for y (II): Distribute 3{x=198−2y594−6y−5y=-22(II): Subtract terms{x=198−2y594−11y=-22(II): LHS−594=RHS−594{x=198−2y-11y=-616(II): LHS/(-11)=RHS/(-11) {x=198−2yy=56 Finally, let's substitute y for 56 in Equation (I). {x=198−2yy=56(I): y=56{x=198−2(56)y=56(I): Multiply{x=198−112y=56(I): Subtract term{x=86y=56 The solution to the system of equations, which is the point of intersection of the lines, is (86,56). In the context of the problem, this means that the measures of the missing angles are 86∘ and 56∘. | |

Exercises 29 The ordered pair (-9,1) is a solution to the given system of equations. Therefore, if we substitute x=-9 and y=1, the equations will still hold true. {a(-9)+b(1)=-31a(-9)−b(1)=-41 ⇔ {-9a+b=-31-9a−b=-41(I)(II) Now we have an system with only a and b as variables. We will solve this system of equations for these variables using the Substitution Method. To do so, we will start by isolating the b-variable in Equation (I). {-9a+b=-31-9a−b=-41(I): LHS+9a=RHS+9a{b=-31+9a-9a−b=-41 Let's now substitute -31+9a for b in Equation (II). {b=-31+9a-9a−b=-41(II): b=-31+9a{b=-31+9a-9a−(-31+9a)=-41 (II): Solve for a (II): Distribute -1{b=-31+9a-9a+31−9a=-41(II): Subtract terms{b=-31+9a-18a+31=-41(II): LHS−31=RHS−31{b=-31+9a-18a=-72(II): LHS/(-18)=RHS/(-18) {b=-31+9aa=4 Finally, we will substitute a=4 in Equation (I), and simplify. {b=-31+9aa=4(I): a=4{b=-31+9(4)a=4(I): Multiply{b=-31+36a=4(I): Add terms{b=5a=4 We found that a=4 and b=5. By substituting these values into the original equations, we obtain a linear system of equations with solution (-9,1). {4x+5y=-314x−5y=-41 | |

Exercises 30 To solve a system of equations using substitution, you start by isolating one of the variables in one of the equations. Then you can substitute whatever that variable equals into the other equation. As an example: {y=2x+15x+2y=3⇒{y=2x+15x+2(2x+1)=3 If we happen to have a system of equations consisting of a vertical line and a horizontal line, we need to think about the general forms for these types of lines.Vertical line: x=h, where h is any real number. Horizontal line: y=k, where k is any real number.Let's suppose that h=1 and k=2. Then we have the following system of equations: {x=1y=2 There is no way to substitute the value of x or y into the other equation. Therefore, we cannot solve this system using substitution and our friend is correct. | |

Exercises 31 This exercise asks us to write a system of linear equations that satisfies the given conditions. To begin, we will make sense of the conditions.Condition #1: Equation 1 must pass through the point (3,-5).Condition #2: Equation 2 must not pass through the point (3,-5). Condition #3: The solution to the system is (-1,7).To satisfy Conditions #1 and #3, Equation 1 must pass through the points (3,-5) and (-1,7). To ensure this happens, we can use the points to determine the slope of the line. Then use either point in point-slope form to write the equation. Since we used both points to determine the slope, it is certain that the line passes through both points. To satisfy Conditions #2 and #3, Equation 2 must pass through (-1,7) and not pass through (3,-5). To ensure this is true we will use the point (-1,7) in point-slope form. Not using the point (3,-5) ensures the line does not pass through that point. As we've ensured that both lines pass through (-1,7), the solution to the system will be (-1,7). To make sure that the system only has one solution, the slopes of the lines must be different. Once we determine the slope of Equation 1, we will choose any different value for the slope of Equation 2.Equation 1 To write Equation 1, we will use point-slope form with either point and the slope of the line. We can find the slope of this line using the slope formula with the points (-1,7) and (3,-5). m=x2−x1y2−y1Substitute (-1,7) & (3,-5)m=3−(-1)-5−7-(-a)=am=3+1-5−7Add termsm=4-5−7\Subtermsm=4−12Calculate quotientm=-3 Thus, the slope of Equation 1 is -3. Now we will write Equation 1 in point-slope form using the slope and either of the points. We will arbitrarily choose to use the point (3,-5). y−y1=m(x−x1)x1=3, y1=-5y−(-5)=m(x−3)-(-a)=ay+5=m(x−3)m=-3y+5=-3(x−3) In point slope form, Equation 1 is y+5=-3(x−3). We can write it in slope intercept form by distributing on the right hand side and isolating y. y+5=-3(x−3)Distribute -3y+5=-3x+9LHS−5=RHS−5y=-3x+4 Equation 1 in slope intercept form is y=-3x+4.Equation 2 We can use point slope form to write Equation 2. We can use any value for the slope that does not equal -3 and we will use the point (-1,7). We will arbitrarily choose to use m=2 for the slope. y−y1=m(x−x1)x1=-1, y1=7y−7=m(x−(-1))-(-a)=ay−7=m(x+1)m=2y−7=2(x+1) In point slope form, Equation 2 is y−7=2(x+1). We can write it in slope intercept form by distributing on the right hand side and isolating y. y−7=2(x+1)Distribute 2y−7=2x+2LHS+7=RHS+7y=2x+9 Equation 2 in slope intercept form is y=2x+9.System The system of equations that satisfies the given conditions is {y=-3x+4y=2x+9 We can graph the system to make sure it is correct. | |

Exercises 32 aThe point of intersection of a system is the x- and y-coordinate where the lines cross. We can determine this point from a graph by looking at the coordinates where the graphs intersect. From the graph, we can see that the lines cross at the point (4,5).bDetermining the solution of a system generally gives an approximate solution. Even if it looks like a point lies on certain coordinates, such as (4,5), it could be that the point actually is (4.0000000000000000001,5). This is so close to (4,5) you would never be able to tell the difference. To verify or to find an exact solution we can solve the system algebraically. It is possible to solve the given system algebraically by using the substitution method.To solve the system {y=x+1y=6−41x. we can substitute x=1 for y in the second equation and solve for x. Once we know x, we can substitute that value into either equation to solve for y. {y=x+1y=6−41x(I)(II)(II): y=x+1{y=x+1x+1=6−41x (II): Solve for x (II): LHS+41x=RHS+41x{y=x+145x+1=6(II): LHS−1=RHS−1{y=x+145x=5(II): LHS⋅4=RHS⋅4{y=x+15x=20(II): LHS/5=RHS/5 {y=x+1x=4 The x coordinate of the solution to the system is 4. To determine the corresponding y value, we can substitute x=4 into the first equation and solve for y. {y=x+1x=4(I): x=4{y=4+1x=4(I): Add terms{y=5x=4 The y coordinate of the solution to the system is 5. From the work above, we have verified that (4,5) is the solution to the system. | |

Exercises 33 Let's solve this exercise by writing and solving a system of equations. To do so, we first need to define the variables. Let p, r, and h be the number of pop, rock, and hip-hop songs, respectively, the radio station plays during a day.Writing the System We are told that a total of 272 songs are played. Therefore, the sum of the three variables must equal 272. p+r+h=272 We are also told that the number of pop songs is 3 times the number of rock songs. Moreover, the number of hip-hop songs is 32 more than the number of rock songs. We can write two more equations using this information. p=3randh=r+32 The three equations we have written form a system of equations. ⎩⎪⎨⎪⎧p+r+h=272p=3rh=r+32(I)(II)(III)Solving the System If we pay close attention to the system, we can see that in Equation (II) and Equation (III) the p- and h-variables are isolated. This means that the most convenient method to solve our system is the Substitution Method. Let's substitute 3r and r+32 for p and h, respectively, in Equation (I). ⎩⎪⎨⎪⎧272=p+r+hp=3rh=r+32(I)(II)(III)(I): p=3r, h=r+32⎩⎪⎨⎪⎧272=3r+r+r+32p=3rh=r+32 (I): Solve for r (I): Add terms⎩⎪⎨⎪⎧272=5r+32p=3rh=r+32(I): LHS−32=RHS−32⎩⎪⎨⎪⎧240=5rp=3rh=r+32(I): LHS/5=RHS/5⎩⎪⎨⎪⎧48=rp=3rh=r+32(I): Rearrange equation ⎩⎪⎨⎪⎧r=48p=3rh=r+32 We found the value of r. Let's now substitute r=48 in Equation (II) and Equation (III). ⎩⎪⎨⎪⎧r=48p=3rh=r+32(II), (III): r=48⎩⎪⎨⎪⎧r=48p=3(48)h=48+32(II): Multiply⎩⎪⎨⎪⎧r=48p=144h=48+32(III): Add terms⎩⎪⎨⎪⎧r=48p=144h=80 The solution to the system is r=48, p=144 and h=80. This means that radio plays 144 pop songs, 48 rock songs and 80 hip-hop songs. | |

Exercises 34 This exercise asks us to write a system of equations to represent the given situation. There are infinitely many solutions to this exercise. What follows is one possibility. Since linear systems have two variables, we can have two different types of coins. We will arbitrarily choose that our coins are quarters and nickels.Let q represent the number of quarters we have. Let n represent the number of nickels we have. The only restraint on our system is that the total amount of money is $2.65. Thus, one equation in the system will be 0.25q+0.05n=2.65. The exercise does not dictate how many of each type of coin we have. We can decide this for ourselves by choosing the number of one type of coins. Suppose we have 10 quarters. That means we have 2.50 in quarters. This means that we must have 0.15 in nickels, or 3 nickels. This results in 13 total coins. We can represent this with the following equation. q+n=13 Combining our equations we get the following system of equations: {0.25q+0.05n=2.65q+n=13 | |

Exercises 35 To find the original number, we will need to form and solve a system of equations. To begin, let's define our variables. Let x represent the number in the tens place and let y represent the number in the ones place. Since we know that the sum of these two numbers is 11, we can write our first equation. x+y=11 Now, let's find the second equation! When the digits are kept in their original places, the value of the unknown number can be written as the expression (10x+y). If we swap their positions, we get the expression (10y+x) and the value increases by 27. As an algebraic equation, this can written as: (10y+x)−(10x+y)=27. Now that we have two equations, we can solve the system using the Substitution Method. {x+y=11(10y+x)−(10x+y)=27(I)(II)(II): Distribute -1{x+y=1110y+x−10x−y=27(II): Subtract term{x+y=119y−9x=27(II): LHS/9=RHS/9{x+y=11y−x=3(II): LHS+x=RHS+x{x+y=11y=x+3 Now that we have y isolated in the second equation, we can use substitution in the first equation to solve for x. {x+y=11y=x+3(I)(II)(I): y=x+3{x+x+3=11y=x+3(I): Add terms{2x+3=11y=x+3(I): LHS−3=RHS−3{2x=8y=x+3(I): LHS/2=RHS/2{x=4y=x+3 We now know that the digit in the tens place is 4. Finally, we substitute this into the second equation to solve for the digit in the ones place. {x=4y=x+3(I)(II)(II): x=4{x=4y=4+3(II): Add terms{x=4y=7 Therefore, the original number was 47. | |

Exercises 36 To find the sum we can combine like terms. We will remove the parenthesis and then group the like terms. (x−4)+(2x−7)Remove parenthesesx−4+2x−7Rearrange termsx+2x−4−7Add and subtract terms3x−11 3x−11 is the sum. | |

Exercises 37 To find the sum we can combine like terms. We will remove the parenthesis and then group the like terms. (5y−12)+(-5y−1)Remove parentheses5y−12−5y−1Rearrange terms5y−5y−12−1Subtract terms-13 -13 is the sum. | |

Exercises 38 To find the difference, we can combine like terms. First we must remove the parentheses. Notice that whenever a parenthesis is preceded by subtraction, we can treat it as though we are distributing -1. (t−8)−(t+15)Distribute -1(t−8)−t−15Remove parenthesest−8−t−15Subtract terms-23 | |

Exercises 39 To find the difference, we can combine like terms. First, we must remove the parentheses. Notice that whenever a parenthesis is preceded by subtraction, we can treat it as though we are distributing -1. (6d+2)−(3d−3)Distribute -1(6d+2)−3d+9Remove parentheses6d+2−3d+9Simplify terms3d+11 | |

Exercises 40 To simplify the expression, we will first use the Distributive Property on both sets of parentheses. Then, we will simplify as much as possible. 4(m+2)+3(6m−4)Distribute 44⋅m+4⋅2+3(6m−4)Distribute 34⋅m+4⋅2+3⋅6m−3⋅4Multiply4m+8+18m−12Rearrange terms4m+18m+8−12Simplify terms22m−4 As 22m and 4 are not like terms, we cannot simplify further. | |

Exercises 41 To simplify the expression, we will first use the Distributive Property on both sets of parentheses. Then, we will simplify as much as possible. 2(5v+6)−6(-9v+2)Distribute 22⋅5v+2⋅6−6(-9v+2)Distribute -62⋅5v+2⋅6+(-6)⋅(-9v)+(-6)⋅2Multiply10v+12+54v−12Rearrange terms10v+54v+12−12Simplify terms64v As there is now only one term, we cannot simplify further. |

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##### 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 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