#### Solving Systems of Linear Equations by Elimination

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###### Exercises
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Exercises 1 To solve a system of equations by addition, we add everything from the LHS and RHS of the first equation to the LHS and RHS of the second equation. This should result in one variable being eliminated, the variable with additive inverse coefficients in the two equations. This is true of the y variables in the following system of equations: {2x+y=13x−y=4​ Let's begin solving this system so that we can see how it works. {2x+y=13x−y=4​(I)(II)​(I):  Add (II){2x+y+3x−y=1+43x−y=4​(I):  Add and subtract terms{5x=53x−y=4​(I):  LHS/5=RHS/5{x=13x−y=4​ We can then continue to solve by substituting x=1 into the second equation. {x=13x−y=4​(II):  x=1{x=13⋅1−y=4​(II):  Multiply{x=13−y=4​(II):  LHS−3=RHS−3{x=1-y=1​(II):  Change signs{x=1y=-1​ The final solution to the system of equations is (1,-1). Any time there are additive inverse coefficients for a variable, Elimination by Addition is often the easiest method.
Exercises 2 We have been given the following system of equations: {2x−3y=-4-5x+9y=7​ and asked to solve by using elimination. To use the elimination method, we need one of the variables to either have the same exact coefficient (so we can subtract to eliminate it) or additive inverse coefficients (so we can add to eliminate it). x−x=0 or x+(-x)=0 Neither of the variables here are like that, so we need to multiply one or both equations by a constant so that our coefficients can be eliminated. {2x−3y=-4-5x+9y=7​(I)(II)​(I):  LHS⋅3=RHS⋅3{6x−9y=-12-5x+9y=7​ Now we have additive inverse coefficients for the y variable. Because they are opposite signs, we will add the equations together to eliminate the y. {6x−9y=-12-5x+9y=7​(I):  Add (II){6x−9y+(-5x+9y)=-12+7-5x+9y=7​(I):  Remove parentheses{6x−9y−5x+9y=-12+7-5x+9y=7​(I):  Add and subtract terms{x=-5-5x+9y=7​ After we have found the solution for one of the variables, in this case x=-5, we can substitute that value into the second equation to solve for the second variable. {x=-5-5x+9y=7​(II):  x=-5{x=-5-5(-5)+9y=7​(II):  -a(-b)=a⋅b{x=-525+9y=7​(II):  LHS−25=RHS−25{x=-59y=-18​(II):  LHS/9=RHS/9{x=-5y=-2​ Finally, we have the final solution to the system of equations: {x=-5y=-2​
Exercises 3 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. {x+2y=13-x+y=5​(I)(II)​​ We can see that the x-terms will eliminate each other if we add (I) to (II). {x+2y=13-x+y=5​(II):  Add (I){x+2y=13-x+y+(x+2y)=5+(13)​ (II): Solve for y (II):  Add and subtract terms{x+2y=133y=18​(II):  LHS/3=RHS/3 {x+2y=13y=6​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {x+2y=13y=6​(I):  y=6{x+2(6)=13y=6​ (I): Solve for x (I):  Multiply{x+12=13y=6​(I):  LHS−12=RHS−12 {x=1y=6​ The solution, or intersection point, of the system of equations is (1,6). 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. {x+2y=13-x+y=5​(I)(II)​(I), (II):  x=1, y=6{(1)+2(6)=13-(1)+(6)=5​(I)(II)​(I), (II):  Multiply{1+12=13-1+6=5​(I)(II)​(I), (II):  Add terms{13=135=5​(I)(II)​ We received two identities, therefore the solution (1,6) is correct.
Exercises 4 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+y=2-4x−y=-17​(I)(II)​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {9x+y=2-4x−y=-17​(II):  Add (I){9x+y=2-4x−y+(9x+y)=-17+(2)​ (II): Solve for x (II):  Add and subtract terms{9x+y=25x=-15​(II):  LHS/5=RHS/5 {9x+y=2x=-3​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {9x+y=2x=-3​(I):  x=-3{9(-3)+y=2x=-3​ (I): Solve for y (I):  Multiply{-27+y=2x=-3​(I):  LHS+27=RHS+27 {y=29x=-3​ The solution, or intersection point, of the system of equations is (-3,29). 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+y=2-4x−y=-17​(I)(II)​(I), (II):  x=-3, y=29{9(-3)+(29)=2-4(-3)−(29)=-17​(I)(II)​(I), (II):  Multiply{-27+29=212−29=-17​(I)(II)​(I), (II):  Add and subtract terms{2=2-17=-17​(I)(II)​ We received two identities, therefore the solution (-3,29) is correct.
Exercises 5 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. {5x+6y=50x−6y=-26​(I)(II)​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {5x+6y=50x−6y=-26​(II):  Add (I){5x+6y=50x−6y+(5x+6y)=-26+(50)​ (II): Solve for x (II):  Add terms{5x+6y=506x=24​(II):  LHS/6=RHS/6 {5x+6y=50x=4​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {5x+6y=50x=4​(I):  x=4{5(4)+6y=50x=4​ (I): Solve for y (I):  Multiply{20+6y=50x=4​(I):  LHS−20=RHS−20{6y=30x=4​(I):  LHS/6=RHS/6 {y=5x=4​ The solution, or intersection point, of the system of equations is (4,5). 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. {5x+6y=50x−6y=-26​(I)(II)​(I), (II):  x=4, y=5{5(4)+6(5)=50(4)−6(5)=-26​(I)(II)​(I), (II):  Multiply{20+30=504−30=-26​(I)(II)​(I), (II):  Add and subtract terms{50=50-26=-26​(I)(II)​ We received two identities, therefore the solution (4,5) is correct.
Exercises 6 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. {-x+y=4x+3y=4​(I)(II)​​ We can see that the x-terms will eliminate each other if we add (II) to (I). {-x+y=4x+3y=4​(I):  Add (II){-x+y+(x+3y)=4+(4)x+3y=4​ (I): Solve for x (I):  Add terms{4y=8x+3y=4​(I):  LHS/4=RHS/4 {y=2x+3y=4​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {y=2x+3y=4​(II):  y=2{y=2x+3(2)=4​ (II): Solve for x (II):  Multiply{y=2x+6=4​(II):  LHS−6=RHS−6 {y=2x=-2​ The solution, or intersection point, of the system of equations is (-2,2). 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. {-x+y=4x+3y=4​(I)(II)​(I), (II):  x=-2, y=2{-(-2)+(2)=4(-2)+3(2)=4​(I)(II)​(I), (II):  Multiply{2+2=4-2+6=4​(I)(II)​(I), (II):  Add terms{4=44=4​(I)(II)​ We received two identities, therefore the solution (-2,2) is correct.
Exercises 7 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. {-3x−5y=-7-4x+5y=14​(I)(II)​​ We can see that the y-terms will eliminate each other if we add (II) to (I). {-3x−5y=-7-4x+5y=14​(I):  Add (II){-3x−5y+(-4x+5y)=-7+(14)-4x+5y=14​ (I): Solve for x (I):  Add and subtract terms{-7x=7-4x+5y=14​(I):  LHS/-7=RHS/-7 {x=-1-4x+5y=14​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {x=-1-4x+5y=14​(II):  x=-1{x=-1-4(-1)+5y=14​ (II): Solve for y (II):  Multiply{x=-14+5y=14​(II):  LHS−4=RHS−4{x=-15y=10​(II):  LHS/5=RHS/5 {x=-1y=2​ The solution, or intersection point, of the system of equations is (-1,2). 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. {-3x−5y=-7-4x+5y=14​(I)(II)​(I), (II):  x=-1, y=2{-3(-1)−5(2)=-7-4(-1)+5(2)=14​(I)(II)​(I), (II):  Multiply{3−10=-74+10=14​(I)(II)​(I), (II):  Add and subtract terms{-7=-714=14​(I)(II)​ We received two identities, therefore the solution (-1,2) is correct.
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. {4x−9y=-21-4x−3y=9​(I)(II)​​ We can see that the x-terms will eliminate each other if we add (I) to (II). {4x−9y=-21-4x−3y=9​(II):  Add (I){4x−9y=-21-4x−3y+(4x−9y)=9+(-21)​ (II): Solve for y (II):  Add and subtract terms{4x−9y=-21-12y=-12​(II):  LHS/-12=RHS/-12 {4x−9y=-21y=1​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {4x−9y=-21y=1​(I):  y=1{4x−9(1)=-21y=1​ (I): Solve for x (I):  Multiply{4x−9=-21y=1​(I):  LHS+9=RHS+9{4x=-12y=1​(I):  LHS/4=RHS/4 {x=-3y=1​ The solution, or intersection point, of the system of equations is (-3,1). 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. {4x−9y=-21-4x−3y=9​(I)(II)​(I), (II):  x=-3, y=1{4(-3)−9(1)=-21-4(-3)−3(1)=9​(I)(II)​(I), (II):  Multiply{-12−9=-2112−3=9​(I)(II)​(I), (II):  Subtract terms{-21=-219=9​(I)(II)​ We received two identities, therefore the solution (-3,1) is correct.
Exercises 9 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. {-y−10=6x5x+y=-10​(I)(II)​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {-y−10=6x5x+y=-10​(II):  Add (I){-y−10=6x5x+y+(-y−10)=-10+(6x)​ (II): Solve for x (II):  Add and subtract terms{-y−10=6x5x−10=6x−10​(II):  LHS+10=RHS+10{-y−10=6x5x=6x​(II):  LHS−5x=RHS−5x{-y−10=6x0=x​(II):  Rearrange equation {-y−10=6xx=0​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {-y−10=6xx=0​(I):  x=0{-y−10=6(0)x=0​ (I): Solve for y (I):  Zero Property of Multiplication{-y−10=0x=0​(I):  LHS+y=RHS+y{-10=yx=0​(I):  Rearrange equation {y=-10x=0​ The solution, or intersection point, of the system of equations is (0,-10). 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. {-y−10=6x5x+y=-10​(I)(II)​(I), (II):  x=0, y=-10{-(-10)−10=6(0)5(0)+(-10)=-10​(I)(II)​(I), (II):  Multiply{10−10=0-10=-10​(I)(II)​(I):  Subtract term{0=0-10=-10​(I)(II)​ We received two identities, therefore the solution (0,-10) is correct.
Exercises 10 We can see that we have 3x in both equations. We can subtract 3x from both sides in the second equation to move it to the left-hand side. Then we can add (I) to (II), eliminate 3x, and solve for y. {3x−30=y7y−6=3x​(I)(II)​(II):  LHS−3x=RHS−3x{3x−30=y7y−6−3x=0​(II):  Add (I){3x−30=y7y−6−3x+(3x−30)=y​ (II):  Solve for y (II):  Remove parentheses{3x−30=y7y−6−3x+3x−30=y​(II):  Add and subtract terms{3x−30=y7y−36=y​(II):  LHS+36=RHS+36{3x−30=y7y=y+36​(II):  LHS−y=RHS−y{3x−30=y6y=36​(II):  LHS/6=RHS/6 {3x−30=yy=6​ Having isolated y, we simply substitute this value into the first equation to calculate the value of x. {3x−30=yy=6​(I):  y=6{3x−30=6y=6​ (I):  Solve for x (I):  LHS+30=RHS+30{3x=36y=6​(I):  LHS/3=RHS/3 {x=12y=6​ To check our solution, we simply substitute its x- and y-coordinates into the first and second equations. If the equations hold true, the solution is correct. {3x−30=y7y−6=3x​(I)(II)​(I), (II):  x=12, y=6{3⋅12−30=?67⋅6−6=?3⋅12​(I), (II):  Multiply{36−30=?642−6=?36​(I), (II):  Subtract term{6=636=36​ Both equations hold true when testing the solution, so it's correct.
Exercises 11 We want to solve the given system and check the obtained solution. Let's do it!Solving the System To solve a system of linear equations using the Elimination Method, one of the variable terms must 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. {x+y=22x+7y=9​(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 Equation (I) by -2, the x-terms will have opposite coefficients. {-2(x+y)=-2(2)2x+7y=9​ ⇒ {-2x−2y=-42x+7y=9​(I)(II)​​ The x-terms will eliminate each other if we add the equations. Let's add Equation (II) to Equation (I). {-2x−2y=-42x+7y=9​(I):  Add (II){-2x−2y+2x+7y=-4+92x+7y=9​(I):  Add terms{5y=52x+7y=9​(I):  LHS/5=RHS/5{y=12x+7y=9​ Now we can solve for x by substituting the value of y into Equation (II) and simplifying. {y=12x+7y=9​(II):  y=1{y=12x+7(1)=9​ (II): Solve for x (II):  Identity Property of Multiplication{y=12x+7=9​(II):  LHS−7=RHS−7{y=12x=2​(II):  LHS/2=RHS/2 {y=1x=1​ The solution to the system, which is the point of intersection of the lines, is (1,1).Checking the Solution To check this solution, we will substitute x=1 and y=1 in the given system and simplify. If doing so results in true statements for both equations in the system, our solution is correct. Otherwise, our solution is incorrect. {x+y=22x+7y=9​(I)(II)​(I), (II):  x=1, y=1{1+1=22(1)+7(1)=9​(II):  Identity Property of Multiplication{1+1=22+7=9​(I), (II):  Add terms⎩⎨⎧​2=2 ✓9=9 ✓​ We obtained two identities. Therefore, the solution (1,1) is correct.
Exercises 12 We want to solve the given system and check the obtained solution. Let's do it!Solving the System To solve a system of linear equations using the Elimination Method, one of the variable terms must 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. {8x−5y=114x−3y=5​(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 Equation (II) by -2, the x-terms will have opposite coefficients. {8x−5y=11-2(4x−3y)=-2(5)​ ⇒ {8x−5y=11-8x+6y=-10​​ The x-terms will eliminate each other if we add the equations. Let's add Equation (I) to Equation (II). {8x−5y=11-8x+6y=-10​(II):  Add (I){8x−5y=11-8x+6y+8x−5y=-10+11​(II):  Add and subtract terms{8x−5y=11y=1​ Now we can solve for x by substituting the value of y into Equation (I) and simplifying. {8x−5y=11y=1​(I):  y=1{8x−5(1)=11y=1​ (I): Solve for y (I):  Identity Property of Multiplication{8x−5=11y=1​(I):  LHS+5=RHS+5{8x=16y=1​(I):  LHS/8=RHS/8 {x=2y=1​ The solution to the system, which is the point of intersection of the lines, is (2,1).Checking the Solution To check this solution, we will substitute x=2 and y=1 in the given system and simplify. If doing so results in true statements for both equations in the system, our solution is correct. Otherwise, our solution is incorrect. {8x−5y=114x−3y=5​(I)(II)​(I), (II):  x=2, y=1{8(2)−5(1)=114(2)−3(1)=5​(I), (II):  Multiply{16−5=118−3=5​(I), (II):  Subtract terms⎩⎨⎧​11=11 ✓5=5 ✓​ We obtained two identities. Therefore, the solution (2,1) is correct.
Exercises 13 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. {11x−20y=283x+4y=36​(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 (II) by 5, the y-terms will have opposite coefficients. {11x−20y=285(3x+4y)=5(36)​ ⇒ {11x−20y=2815x+20y=180​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {11x−20y=2815x+20y=180​(II):  Add (I){11x−20y=2815x+20y+(11x−20y)=180+(28)​ (II): Solve for x (II):  Add and subtract terms{11x−20y=2826x=208​(II):  LHS/26=RHS/26 {11x−20y=28x=8​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {11x−20y=28x=8​(I):  x=8{11(8)−20y=28x=8​ (I): Solve for y (I):  Multiply{88−20y=28x=8​(I):  LHS−88=RHS−88{-20y=-60x=8​(I):  LHS/-20=RHS/-20 {y=3x=8​ The solution, or intersection point, of the system of equations is (8,3). 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. {11x−20y=283x+4y=36​(I)(II)​(I), (II):  x=8, y=3{11(8)−20(3)=283(8)+4(3)=36​(I)(II)​(I), (II):  Multiply{88−60=2824+12=36​(I)(II)​(I), (II):  Add and subtract terms{28=2836=36​(I)(II)​ We received two identities, therefore the solution (8,3) is correct.
Exercises 14 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. {10x−9y=46-2x+3y=10​(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 (II) by 3, the y-terms will have opposite coefficients. {10x−9y=463(-2x+3y)=3(10)​ ⇒ {10x−9y=46-6x+9y=30​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {10x−9y=46-6x+9y=30​(II):  Add (I){10x−9y=46-6x+9y+(10x−9y)=30+(46)​ (II): Solve for x (II):  Add and subtract terms{10x−9y=464x=76​(II):  LHS/4=RHS/4 {10x−9y=46x=19​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {10x−9y=46x=19​(I):  x=19{10(19)−9y=46x=19​ (I): Solve for y (I):  Multiply{190−9y=46x=19​(I):  LHS−190=RHS−190{-9y=-144x=19​(I):  LHS/-9=RHS/-9 {y=16x=19​ The solution, or intersection point, of the system of equations is (19,16). 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. {10x−9y=46-2x+3y=10​(I)(II)​(I), (II):  x=19, y=16{10(19)−9(16)=46-2(19)+3(16)=10​(I)(II)​(I), (II):  Multiply{190−144=46-38+48=10​(I)(II)​(I), (II):  Add and subtract terms{46=4610=10​(I)(II)​ We received two identities, therefore the solution (19,16) is correct.
Exercises 15 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. {4x−3y=85x−2y=-11​(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 -2 and multiply (II) by 3, the y-terms will have opposite coefficients. {-2(4x−3y)=-2(8)3(5x−2y)=3(-11)​ ⇒ {-8x+6y=-1615x−6y=-33​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {-8x+6y=-1615x−6y=-33​(II):  Add (I){-8x+6y=-1615x−6y+(-8x+6y)=-33+(-16)​ (II): Solve for x (II):  Add and subtract terms{-8x+6y=-167x=-49​(II):  LHS/7=RHS/7 {-8x+6y=-16x=-7​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {-8x+6y=-16x=-7​(I):  x=-7{-8(-7)+6y=-16x=-7​ (I): Solve for y (I):  Multiply{56+6y=-16x=-7​(I):  LHS−56=RHS−56{6y=-72x=-7​(I):  LHS/6=RHS/6 {y=-12x=-7​ The solution, or intersection point, of the system of equations is (-7,-12). 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. {4x−3y=85x−2y=-11​(I)(II)​(I), (II):  x=-7, y=-12{4(-7)−3(-12)=85(-7)−2(-12)=-11​(I)(II)​(I), (II):  Multiply{-28+36=835+24=-11​(I)(II)​(I), (II):  Add terms{8=8-11=-11​(I)(II)​ We received two identities, therefore the solution (-7,-12) is correct.
Exercises 16 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. {-2x−5y=93x+11y=4​(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 2, the x-terms will have opposite coefficients. {3(-2x−5y)=3(9)2(3x+11y)=2(4)​ ⇒ {-6x−15y=276x+22y=8​​ We can see that the x-terms will eliminate each other if we add (I) to (II). {-6x−15y=276x+22y=8​(II):  Add (I){-6x−15y=276x+22y+(-6x−15y)=8+(27)​ (II): Solve for x (II):  Add and subtract terms{-6x−15y=277y=35​(II):  LHS/7=RHS/7 {-6x−15y=27y=5​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {-6x−15y=27y=5​(I):  y=5{-6x−15(5)=27y=5​ (I): Solve for y (I):  Multiply{-6x−75=27y=5​(I):  LHS+75=RHS+75{-6x=102y=5​(I):  LHS/-6=RHS/-6 {x=-17y=5​ The solution, or intersection point, of the system of equations is (-17,5). 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. {-2x−5y=93x+11y=4​(I)(II)​(I), (II):  x=-17, y=5{-2(-17)−5(5)=93(-17)+11(5)=4​(I)(II)​(I), (II):  Multiply{34−25=9-51+55=4​(I)(II)​(I), (II):  Add and subtract terms{9=94=4​(I)(II)​ We received two identities, therefore the solution (-17,5) is correct.
Exercises 17 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=396x+13y=-9​(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 2 and multiply (II) by -3, the x-terms will have opposite coefficients. {2(9x+2y)=2(39)-3(6x+13y)=-3(-9)​ ⇒ {18x+4y=78-18x−39y=27​​ We can see that the x-terms will eliminate each other if we add (I) to (II). {18x+4y=78-18x−39y=27​(II):  Add (I){18x+4y=78-18x−39y+(18x+4y)=27+(78)​ (II): Solve for y (II):  Add and subtract terms{18x+4y=78-35y=105​(II):  LHS/-35=RHS/-35 {18x+4y=78y=-3​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {18x+4y=78y=-3​(I):  y=-3{18x+4(-3)=78y=-3​ (I): Solve for x (I):  Multiply{18x−12=78y=-3​(I):  LHS+12=RHS+12{18x=90y=-3​(I):  LHS/18=RHS/18 {x=5y=-3​ The solution, or intersection point, of the system of equations is (5,-3). 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=396x+13y=-9​(I)(II)​(I), (II):  x=5, y=-3{9(5)+2(-3)=396(5)+13(-3)=-9​(I)(II)​(I), (II):  Multiply{45−6=3930−39=-9​(I)(II)​(I), (II):  Subtract term{39=39-9=-9​(I)(II)​ We received two identities, therefore the solution (5,-3) is correct.
Exercises 18 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. {12x−7y=-28x+11y=30​(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 -2 and multiply (II) by 3, the x-terms will have opposite coefficients. {-2(12x−7y)=-2(-2)3(8x+11y)=3(30)​ ⇒ {-24x+14y=424x+33y=90​​ We can see that the x-terms will eliminate each other if we add (I) to (II). {-24x+14y=424x+33y=90​(II):  Add (I){-24x+14y=424x+33y+(-24x+14y)=90+(4)​ (II): Solve for x (II):  Add and subtract terms{-24x+14y=447y=94​(II):  LHS/47=RHS/47 {-24x+14y=4y=2​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {-24x+14y=4y=2​(I):  y=2{-24x+14(2)=4y=2​ (I): Solve for y (I):  Multiply{-24x+28=4y=2​(I):  LHS−28=RHS−28{-24x=-24y=2​(I):  LHS/-24=RHS/-24 {x=1y=2​ The solution, or intersection point, of the system of equations is (1,2). 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. {12x−7y=-28x+11y=30​(I)(II)​(I), (II):  x=1, y=2{12(1)−7(2)=-28(1)+11(2)=30​(I)(II)​(I), (II):  Multiply{12−14=-28+22=30​(I)(II)​(I), (II):  Add and subtract terms{-2=-230=30​(I)(II)​ We received two identities, therefore the solution (1,2) is correct.
Exercises 19 It seems that the first equation is added to the second one. However, the sum of the x terms is incorrect as we have calculated 5x−x instead of 5x+x. Let's correct this mistake and solve the system. {5x−7y=16x+7y=8​(I)(II)​(I):  Add (II){5x−7y+(x+7y)=16+8x+7y=8​(I):  Remove parentheses{5x−7y+x+7y=16+8x+7y=8​(I):  Add and subtract terms{6x=24x+7y=8​(I):  LHS/6=RHS/6{x=4x+7y=8​
Exercises 20 When we multiply the second equation by -4, we have to multiply both sides. In the solution given, the right-hand side of the second equation was not multiplied by -4. {4x+3y=8x−2y=-13​(I)(II)​(II):  LHS⋅-4=RHS⋅-4{4x+3y=8-4x+8y=52​(II):  Add I{4x+3y=8-4x+8y+4x+3y=52+8​(II):  Add terms{4x+3y=811y=60​(II):  LHS/11=RHS/11⎩⎨⎧​4x+3y=8y=1160​​
Exercises 21 Let's formulate the total cost of service in this center. If we let q represent the quarts of oil used, we have the following equation to describe the total cost: x+yq, where x and y are as described in the exercise. Customer A used 5 quarts of oil and paid \$22.45 in total. Using the expression above, we can write this as the equation: x+5y=22.45. Similarly, given the information about customer B, we have the equation: x+7y=25.45. Combining these, we get a system of linear equations which we can solve using the method of elimination. {x+5y=22.45x+7y=25.45​(I)(II)​(II):  Subtract (I){x+5y=22.45x+7y−(x+5y)=25.45−22.45​ (II):  Solve for y (II):  Remove parentheses and change signs{x+5y=22.45x+7y−x−5y=25.45−22.45​(II):  Subtract terms{x+5y=22.452y=3​(II):  LHS/2=RHS/2 {x+5y=22.45y=1.50​ Therefore, the cost per quart is \$1.50. To find the fixed fee for oil change x, we can substitute the value of y into the first equation. {x+5y=22.45y=1.50​(I):  y=1.50{x+5⋅1.50=22.45y=1.50​ (I):  Solve for x (I):  Multiply{x+7.5=22.45y=1.50​(I):  LHS−7.5=RHS−7.5 {x=14.95y=1.50​
Exercises 22 We will write an equation for each person and combine them into a system of equations. Since price per song is represented by x and price per album is represented by y, all we have to do is to multiply the number of songs and the number of albums by respective prices. The sum of these products equated with the amount of money each person pays will form equations. Let's start with the first person. 6x+2y=25.92​ Now, we will form an equation for the second person. 4x+3y=33.93​ Using two equations above, we can form a system of equations and solve it for x and y. Let's isolate y in the first equation to solve the system using elimination method. {6x+2y=25.924x+3y=33.93​(I)(II)​(I): LHS/2=RHS/2{3x+y=12.964x+3y=33.93​(I): LHS−3x=RHS−3x{y=12.96−3x4x+3y=33.93​(II): y=12.96−3x{y=12.96−3x4x+3(12.96−3x)=33.93​(II): Distribute 3{y=12.96−3x4x+38.88−9x=33.93​(II): LHS−38.88=RHS−38.88{y=12.96−3x4x−9x=4.95​(II): Subtract term{y=12.96−3x-5x=-4.95​(II): LHS/-5=RHS/-5{y=12.96−3xx=0.99​ Now that we have x=0.99 we can substitute it into the first equation to solve it for y. {y=12.96−3xx=0.99​(I): x=0.99{y=12.96−3(0.99)x=0.99​(I): Multiply{y=12.96−2.97x=0.99​(I): Subtract term{y=9.99x=0.99​ The solution to the system is x=0.99 and y=9.99. In the context of the word problem it means that the website charges \$9.99 for an album and \$0.99 for a song.
Exercises 23 If one of the variable terms is eliminated when one equation is added to or subtracted from the other equation, you can use Elimination Method to solve a system of linear equations. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. {3x+2y=42y=8−5x​(I)(II)​​ We can see that the y-terms in this system will eliminate each other if we subtract (I) from (II). {3x+2y=42y=8−5x​(I)(II)​(II):  Subtract (I){3x+2y=42y−(3x+2y)=8−5x−4​ (II): Solve for m (II):  Remove parentheses and change signs{3x+2y=42y−3x−2y=8−5x−4​(II):  Add and subtract terms{3x+2y=4−3x−=4−5x​(II):  LHS+5x=RHS+5x{3x+2y=42x=4​(II): LHS/2=RHS/2 {3x+2y=4x=2​ With this, we can now solve for y by substituting the value of x into either equation and simplifying. {3x+2y=4x=2​(I):  x=2{3(2)+2y=4x=2​ (I): Solve for p (I):  Multiply{6+2y=4x=2​(I):  LHS−6=RHS−6{2y=-2x=2​(I):  LHS/2=RHS/2 {y=-1x=2​ The solution to the system of equations, the point of intersection, is (2,-1).
Exercises 24 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! {-6y+2=-4xy−2=x​(I)(II)​(I): x=y−2{-6y+2=-4(y−2)y−2=x​(I): Distribute -4{-6y+2=-4y+8y−2=x​(I): LHS+4y−2=RHS+4y−2{-2y=6y−2=x​(I):  LHS/-2=RHS/-2{y=-3y−2=x​ 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. {y=-3y−2=x​(II): y=-3{y=-3-3−2=x​(II):  Subtract term{y=-3-5=x​(II): Rearrange equation{y=-3x=-5​ The solution, or point of intersection, to this system of equations is the point (-5,-3).
Exercises 25 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! ⎩⎨⎧​y−x=2y=-41​x+7​(I)(II)​(I): y=-41​x+7⎩⎪⎨⎪⎧​-41​x+7−x=2y=-41​x+7​(I): LHS−7=RHS−7⎩⎪⎨⎪⎧​-41​x−x=-5y=-41​x+7​(I): Subtract term⎩⎪⎨⎪⎧​-45​x=-5y=-41​x+7​(I):  LHS/-45​=RHS/-45​⎩⎨⎧​x=4y=-41​x+7​ Great! Now, to find the value of y, we need to substitute x=4 into either one of the equations in the given system. Let's use the first equation. ⎩⎨⎧​x=4y=-41​x+7​(I)(II)​(I): x=4⎩⎨⎧​x=4y=-41​(4)+7​(I)(II)​(I):  Multiply{x=4y=-1+7​(I): Add terms{x=4y=6​ The solution, or point of intersection, to this system of equations is the point (4,6).
Exercises 26 Since the first equation has a variable with a coefficient of 1, the substitution method might be the easiest. ⎩⎪⎨⎪⎧​3x+y=31​2x−3y=38​​(I)(II)​ ​ We will isolate y in the first equation and substitute it in the second equation. ⎩⎪⎨⎪⎧​3x+y=31​2x−3y=38​​(I)(II)​(I): LHS−3x=RHS−3x⎩⎪⎨⎪⎧​y=31​−3x2x−3y=38​​(I)(II)​(II): y=31​−3x⎩⎪⎨⎪⎧​y=31​−3x2x−3(31​−3x)=38​​(I)(II)​(II): Distribute -3⎩⎪⎨⎪⎧​y=31​−3x2x−1+9x=38​​(I)(II)​(II): LHS+1=RHS+1⎩⎪⎨⎪⎧​y=31​−3x2x+9x=38​+1​(I)(II)​(II): Rewrite 1 as 33​⎩⎪⎨⎪⎧​y=31​−3x2x+9x=38​+33​​(I)(II)​(II): Add terms⎩⎪⎨⎪⎧​y=31​−3x11x=311​​(I)(II)​(II): LHS/11=RHS/11⎩⎪⎨⎪⎧​y=31​−3xx=31​​(I)(II)​ Now, we will substitute x=31​ in the first equation and solve for y. ⎩⎪⎨⎪⎧​y=31​−3xx=31​​(I)(II)​(I): x=31​⎩⎪⎨⎪⎧​y=31​−3(31​)x=31​​(I)(II)​(I): Multiply⎩⎪⎨⎪⎧​y=31​−1x=31​​(I)(II)​(I): Subtract term⎩⎪⎨⎪⎧​y=-32​x=31​​(I)(II)​ The solution, or intersection point, of the system of equations is (31​,-32​).
Exercises 27 To use the elimination method without needing to multiply an equation first, one of the variables should either have the same coefficient or additive opposite coefficients in the two equations. We need the coefficients to either add or subtract to be 0. The given equations are {ax+3y=24x+5y=6​ Notice that the x-term in the second equation is 4x. This means the x-term in the first equation must be either -4 or 4: {-4x+3y=24x+5y=6​or{4x+3y=24x+5y=6​ To cancel out the x-variables in the first system we have to add the equations. To cancel them out in the second system we have to subtract.
Exercises 28 a. To determine the number of students who chose breakfast and lunch, we can subtract the 25 students who chose dinner from the total 50 students who were surveyed. Since 50−25=25, 25 students chose breakfast and lunch. b. To write a system of equations, we should first define variables. Let b be the number of students who chose breakfast, and let l be the number of students who chose breakfast.Equation 1: In Part A, we found that 25 total students chose breakfast and lunch. Thus, b+l=25.Equation 2: It is given that the number of students who chose lunch is five more than the number of students who chose breakfast. Thus, we can writel=b+5. The system we've created is {b+l=25l=b+5​(I)(II)​ c. To solve the system we created in Part B, we could use the substitution method. Substituting Equation II into Equation I will allow us to solve for b. Then, we'd solve for l by substituting b into Equation II. If we add the solutions of b and l, we would find the number of students who chose breakfast and lunch.
Exercises 29 Both the substitution method and the elimination method for solving systems are useful strategies. Sometimes one method will take fewer steps than the other. However, it is not always the case that substitution will take fewer steps. Suppose we have the system {8x−5y=114x−3y=5​(I)(II)​ We will solve the system using both methods.Solving with Substitution To solve using substitution, we must first isolate one variable in one equation. We will arbitrarily choose to isolate y in Equation II. {8x−5y=114x−3y=5​(I)(II)​ (II):  Solve for y (II): LHS−4x=RHS−4x{8x−5y=11-3y=-4x+5​(II):  Change signs{8x−5y=113y=4x−5​(II):  LHS/3=RHS/3 {8x−5y=11y=34x−5​​ Having isolated y, we can substitute (II) into (I) and solve for x. {8x−5y=11y=34x−5​​(I): y=34x−5​{8x−5⋅34x−5​=11y=34x−5​​ (I):  Solve for x (I):  LHS⋅3=RHS⋅3{24x−5(4x−5)=33y=34x−5​​(I):  Distribute -5{24x−20x+25=33y=34x−5​​(I): Subtract terms{4x+25=33y=34x−5​​(I): LHS−25=RHS−25{4x=8y=34x−5​​(I): LHS/4=RHS/4 {x=2y=34x−5​​ By substituting x=2 into (II), we can solve for y. {x=2y=34x−5​​(II): x=2{x=2y=34⋅2−5​​ (II):  Simplify RHS (II): Multiply{x=2y=38−5​​(II): Subtract term{x=2y=33​​(II): Calculate quotient {x=2y=1​ The solution to the system is (2,1). From the work above, it is clear that using the substitution method to solve this system was extremely convoluted involving lots of fractions and opportunities to mess up.Solving with Elimination To solve the system with elimination, one of the variables should either have the same coefficient or additive opposite coefficients in the two equations. Recall the original system from above. {8x−5y=114x−3y=5​(I)(II)​. Notice that the coefficients of the x terms are 8 and 4. If we multiply the second equation by 2, the coefficients will be 8 and 8. Then we would be able to use the elimination method by subtracting. {8x−5y=114x−3y=5​(I)(II)​(II): LHS⋅2=RHS⋅2{8x−5y=118x−6y=10​(II): Subtract (I){8x−5y=118x−6y−(8x−5y)=10−11​ (II):  Solve for y (II):  Remove parentheses and change signs{8x−5y=118x−6y−8x+5y=10−11​(II):  Add and subtract terms{8x−5y=11-y=-1​(II):  Change signs {8x−5y=11y=1​ We can now use the value y=1 to solve for x by substituting it into Equation (I) and solving for x. {8x−5y=11y=1​(I): y=1{8x−5⋅1=11y=1​ (I):  Solve for x (I): Multiply{8x−5=11y=1​(I): LHS+5=RHS+5{8x=16y=1​(I): LHS/8=RHS/8 {x=2y=1​ The solution to the system is (2,1).Conclusion From the work above, it is clear that the elimination method yielded fewer steps for the example provided. Thus, substitution does not always take fewer steps than elimination. Our friend is incorrect.
Exercises 31 To begin, we will make sense of the given information.Original Rectangle: the dimensions of this rectangle are l and w, which represent the length and the width, respectively. The perimeter of this rectangle is 18 inches. New rectangle: the dimensions of this rectangle are 3l and 2w, which represent the length and the width, respectively. The perimeter of this rectangle is 46 inches. Writing the system We can write one equation that represents the perimeter of each rectangle. To find the perimeter, we add all the sides. This gives the following equations.Original Rectangle: 2l+2w=18New Rectangle: 2(3l)+2(2w)=46→6l+4w=46Thus, the system {2l+2w=186l+4w=46​(I)(II)​ can be used to determine the length and width of the original rectangle.Solving the system We will use the elimination method to solve the system. If we multiply Equation I by -2, the w terms will cancel out when we add the equations. From here, we will be able to solve for l. {2l+2w=186l+4w=46​(I)(II)​(I): LHS⋅-2=RHS⋅-2{-4l−4w=-366l+4w=46​(I): Add (II){-4l−4w+(6l+4w)=-36+(46)6l+4w=46​(I): Remove parentheses{-4l−4w+6l+4w=-36+466l+4w=46​(I): Rearrange terms{-4l+6l−4w+4w=-36+466l+4w=46​(I): Add terms{2l=106l+4w=46​(I): LHS/1=RHS/1{l=56l+4w=46​ The length of the original rectangle is 5 inches. We can use l=5 to determine w by substituting it into Equation II. {l=56l+4w=46​(II): l=5{l=56(5)+4w=46​(II): Multiply{l=530+4w=46​(II): LHS−30=RHS−30{l=54w=16​(II): LHS/4=RHS/4{l=5w=4​ The width of the original rectangle is 4 inches. b. Recall that the length and width of the new rectangle are 3l and 2w. We can calculate the exact value of both by substituting the values of l and w.New width: 2w=2(4)=8 inchesNew length: 3l=3(5)=15 inches
Exercises 32 If we can rewrite System 1 as System 2, we know that they are equivalent and have the same solution. Since the first equation in System 1 only has an x variable, we can safely assume that the y-variable has been eliminated. Either it's been eliminated through substitution or elimination. Let's eliminate the y-variable in the first equation of system 1, isolate x, and see where we end up. {3x−2y=8x+y=6​(I)(II)​(II):  LHS⋅2=RHS⋅2{3x−2y=82x+2y=12​(I): Add (II){3x−2y+(2x+2y)=8+122x+2y=12​(I):  Remove parentheses{3x−2y+2x+2y=8+122x+2y=12​(I): Add and subtract terms{5x=202x+2y=12​(II):  LHS/2=RHS/2{5x=20x+y=6​ There we are! Since we were able to rewrite System 1 as System 2, we know they are equivalent.
Exercises 33 We are told that you are making 6 quarts of fruit punch. If we let x represent the volume of 20% juice you mix and y represent the volume of 100% juice you mix, we can write the first equation. x+y=6​ We are also told that you want your mixture to contain 80% fruit juice. If we multiply each volume from the first equation by its fruit juice percentage, we get the second equation. Note that to write the second equation we have to convert each percentage into decimal. 100%80%20%​⇒⇒⇒​10.80.2​ Let's write the second equation using these converted values. 0.2x+y=0.8⋅6⇒0.2x+y=4.8​ Combining these equations, we can write them as a system. {x+y=60.2x+y=4.8​(I)(II)​​ Since y-coefficients are both equal to 1, we will use elimination method to solve the system. {x+y=60.2x+y=0.8⋅6​(I)(II)​(II): Subtract I{x+y=60.2x+y−(x+y)=4.8−6​(I)(II)​(II): Distribute -1{x+y=60.2x+y−x−y=4.8−6​(I)(II)​(II): Subtract terms{x+y=6-0.8x=-1.2​(I)(II)​Multiply by -45​{x+y=6x=1.5​(I)(II)​ Now, let's substitute x=1.5 in the first equation. {x+y=6x=1.5​(I)(II)​(I): x=1.5{1.5+y=6x=1.5​(I)(II)​(I): LHS−1.5=RHS−1.5{y=4.5x=1.5​(I)(II)​ The solution, or intersection point, of the system of equations is (1.5,4.5). In the context of the word problem, it means that you should mix 1.5 quarts of 20% juice and 4.5 quarts of 100% juice.
Exercises 34 First, let b represent the speed of the boat and c the speed of the current. Then we can write a distance formula for each trip. d=r⋅t​ Before we substitute values into the formula, we need a graph to present all the given information. Note that we have converted 60 minutes into 1 hour and 40 minutes into 32​ hour.While traveling downstream the rate equals the sum of boat's speed and current's speed. We are told that the trip takes 32​ hour. To write the first equation we multiply rate by time and equate to the distance of 20 miles. 20=(b+c)32​​ Analogically, while traveling upstream the rate equals the difference between boat's speed and current's speed. We know that the trip takes 1 hour. To write the second equation we multiply rate by time and equate to the distance of 20 miles. 20=(b−c)1​ Let's combine created equations into a system. ⎩⎨⎧​20=(b+c)32​20=b−c​(I)(II)​⇒⎩⎨⎧​20=32​b+32​c20=b−c​​ If we multiply the first equation by 23​, we have both b-coefficients equal to 1 and can use elimination method. ⎩⎨⎧​20=32​b+32​c20=60b−60c​(I)(II)​(I): Multiply by 23​{30=b+c20=b−c​(I): Subtract II{30−20=b+c−(b−c)20=b−c​(I): -(b−a)=a−b{30−20=b+c+c−b20=b−c​(I): Add and subtract terms{10=2c20=b−c​(I): LHS/12=RHS/12{5=c20=b−c​ Now, let's substitute c=5 in the second equation. {c=520=b−c​(II): c=5{c=520=b−5​(II): Multiply{c=520=b−5​(II): LHS+5=RHS+5{c=525=b​(II): Rearrange equation{c=5b=25​ The solution, or intersection point, of the system of equations is b=25 and c=5. In the context of the word problem, this means that the speed of the stream is 5 miles per hour.
Exercises 35 Notice that, in the given system, there are three equations and three variables. This means that the solution to the system will take the form (x,y,z). We will begin by isolating y in Equation (III). ⎩⎪⎨⎪⎧​x+7y+3z=293z+x−2y=-75y=10−2x​(I)(II)(III)​(III): LHS/5=RHS/5⎩⎪⎨⎪⎧​x+7y+3z=293z+x−2y=-7y=2−0.4x​(I): y=2−0.4x⎩⎪⎨⎪⎧​x+7(2−0.4x)+3z=293z+x−2(2−0.4x)=-7y=2−0.4x​ (I), (II):  Simplify (I): Distribute 7⎩⎪⎨⎪⎧​x+14−2.8x+3z=293z+x−2(2−0.4x)=-7y=2−0.4x​(I): Distribute -2⎩⎪⎨⎪⎧​x+14−2.8x+3z=293z+x−4+0.8x=-7y=2−0.4x​(I), (II): Add and subtract terms⎩⎪⎨⎪⎧​14−1.8x+3z=293z−4+1.8x=-7y=2−0.4x​(I): LHS−14=RHS−14⎩⎪⎨⎪⎧​-1.8x+3z=153z−4+1.8x=-7y=2−0.4x​(II):  LHS+4=RHS+4⎩⎪⎨⎪⎧​-1.8x+3z=153z+1.8x=-3y=2−0.4x​(II):  Rearrange terms ⎩⎪⎨⎪⎧​-1.8x+3z=151.8x+3z=-3y=2−0.4x​ Notice that the coefficients of the x terms in Equation (I) and Equation (II) are opposite. Thus, if we add these equations together, we can eliminate x and solve for z. ⎩⎪⎨⎪⎧​-1.8x+3z=151.8x+3z=-3y=2−0.4x​(II): Add (I)⎩⎪⎨⎪⎧​-1.8x+3z=151.8x+3z+(-1.8x+3z)=-3+15y=2−0.4x​ (II):  Solve for z (II): Remove parentheses⎩⎪⎨⎪⎧​-1.8x+3z=151.8x+3z−1.8x+3z=-3+15y=2−0.4x​(II): Add terms⎩⎪⎨⎪⎧​-1.8x+3z=156z=12y=2−0.4x​(II): LHS/6=RHS/6 ⎩⎪⎨⎪⎧​-1.8x+3z=15z=2y=2−0.4x​ The value z=2 is the z-coordinate of the solution. Let's substitute this into Equation (I) and solve for x. ⎩⎪⎨⎪⎧​-1.8x+3z=15z=2y=2−0.4x​(I): z=2⎩⎪⎨⎪⎧​-1.8x+3⋅2=15z=2y=2−0.4x​ (I):  Solve for x (I): Multiply⎩⎪⎨⎪⎧​-1.8x+6=15z=2y=2−0.4x​(I): LHS−6=RHS−6⎩⎪⎨⎪⎧​-1.8x=9z=2y=2−0.4x​(I): LHS/(-1.8)=RHS/(-1.8) ⎩⎪⎨⎪⎧​x=-5z=2y=2−0.4x​ The value x=-5 is the x-coordinate of the solution. We can substitute this into Equation (III) to solve for y. ⎩⎪⎨⎪⎧​x=-5z=2y=2−0.4x​(III): x=-5⎩⎪⎨⎪⎧​x=-5z=2y=2−0.4(-5)​ (III):  Solve for y (III):  -a(-b)=a⋅b⎩⎪⎨⎪⎧​x=-5z=2y=2+2​(III): Add terms ⎩⎪⎨⎪⎧​x=-5z=2y=4​ The solution to the system is (-5,2,4).
Exercises 36 To solve an equation with a variable on both sides, we gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. 5d−8=1+5dLHS−5d=RHS−5d5d−8−5d=1+5d−5dSubtract term-8≠1 Simplifying the equation resulted in a contradiction. Thus, the equation has no solution.
Exercises 37 To solve an equation with a variable on both sides, we gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. 9+4t=12−4tLHS+4t=RHS+4t9+4t+4t=12−4t+4tAdd terms9+8t=12LHS−9=RHS−99+8t−9=12−9Subtract term8t=3LHS/8=RHS/8t=83​ The solution to the equation is t=83​.
Exercises 38 To solve an equation with a variable on both sides, we gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. Before we do that, we have a product that can be simplified using the Distributive Property. 3n+2=2(n−3)Distribute 23n+2=2n−6 Now we can continue to solve using the Properties of Equality. 3n+2=2n−6LHS−2n=RHS−2n3n+2−2n=2n−6−2nSubtract termn+2=-6LHS−2=RHS−2n+2−2=-6−2Subtract termn=-8 The solution to the equation is n=-8.
Exercises 39 To solve an equation with a variable on both sides, we gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality. Before we do that, we have a product that can be simplified using the Distributive Property. -3(4−2v)=6v−12Distribute -3-12+6v=6v−12 Now we can continue to solve using the Properties of Equality. -12+6v=6v−12LHS−6v=RHS−6v-12+6v−6v=6v−12−6vSubtract term-12=-12 Simplifying the equation resulted in an identity. Thus, the equation has infinitely many solutions.
Exercises 40 When lines are parallel, they have the same slope. y=-2x+7​ Because of this, we know that all lines that are parallel to our given line will have a slope of -2. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=-2x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (4,-1). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=-2x+bx=4, y=-1-1=-2(4)+b Solve for b (-a)b=-ab-1=-8+bLHS+8=RHS+87=bRearrange equation b=7 Now that we have the y-intercept, we can see that this is the given equation. y=-2x+7​
Exercises 41 When lines are parallel, they have the same slope. y=5x−3​ Because of this, we know that all lines that are parallel to our given line will have a slope of 5. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=5x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (0,6). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=5x+bx=0, y=66=5(0)+bZero Property of Multiplication6=bRearrange equationb=6 Now that we have the y-intercept, we can make an equation for the line. y=5x+6​ The line given by the equation above is parallel to y=5x−3 and passes through (0,6).
Exercises 42 When lines are parallel, they have the same slope. y=32​x+1​ Because of this, we know that all lines that are parallel to our given line will have a slope of 32​. This means we can write a general equation in slope-intercept form for all lines parallel to the given equation. y=32​x+b​ We are asked to write the equation of a line parallel to the given equation that passes through the given point (-5,-2). By substituting this point into the general equation for x and y, we will be able to solve for the y-intercept b of the parallel line. y=32​x+bx=-5, y=-2-2=32​(-5)+b Solve for b a(-b)=-a⋅b-2=-310​+bLHS+310​=RHS+310​-2+310​=ba=22⋅a​-24​+310​=bba​=b⋅3a⋅3​-612​+310​=bba​=b⋅2a⋅2​-612​+620​=bAdd fractions68​=bba​=b/2a/2​34​=bRearrange equation b=34​ Now that we have the y-intercept, we can make an equation for the line. y=32​x+34​​ The line given by the equation above is parallel to y=32​x+1 and passes through point (-5,-2).
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