3 Variable Systems of Equations – Elimination Method

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Question 1 of 5

Solve the following systems of equations by elimination.

3 x - y + 2 z = 5

$3x-y+2z=5$

4 x + 2 y - z = 6

$4x+2y-z=6$

5 x - 3 y + z = 1

$5x-3y+z=1$

$x =$ $x=$ (1)

$y =$ $y=$ (2)

$z =$ $z=$ (2)

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Elimination Method

$1)$ $1)$ make sure 2 equations have a variable with same coefficients but opposite signs
$2)$ $2)$ add or subtract the equations so that one variable is cancelled
$3)$ $3)$ solve for the variable that remains
$4)$ $4)$ substitute known value to one of the equations to solve for the other variable

First, label the equations

1

$1$ ,

2

$2$ and

3

$3$ respectively.

$3 x - y + 2 z$ $3x-y+2z$	$=$ $=$	$5$ $5$	Equation $1$ $1$
$4 x + 2 y - z$ $4x+2y-z$	$=$ $=$	$6$ $6$	Equation $2$ $2$
$5 x - 3 y + z$ $5x-3y+z$	$=$ $=$	$1$ $1$	Equation $3$ $3$

Next, add equation

2

$2$ to equation

3

$3$ since their variable

z

$z$ both have

1

$1$ as coefficient but with opposite signs. Label the result as equation

A

$A$ .

$4 x + 2 y - z$ $4x+2y-z$	$=$ $=$	$6$ $6$
$+$ $+$ $5 x - 3 y + z$ $5x-3y+z$	$=$ $=$	$1$ $1$	$- z + z$ $-z+z$ cancels out

$9 x - y$ $9x-y$	$=$ $=$	$7$ $7$	Equation $A$ $A$

Next, multiply the values of equation

2

$2$ by

2

$2$ and add the product to equation

1

$1$ . Label the result as equation

B

$B$ .

$2 \times$ $2xx$ $(4 x + 2 y - z)$ $(4x+2y-z)$	$=$ $=$	$2 \times$ $2xx$ $6$ $6$	Multiply equation $2$ $2$ by $2$ $2$
$8 x + 4 y - 2 z$ $8x+4y-2z$	$=$ $=$	$12$ $12$

$8 x + 4 y - 2 z$ $8x+4y-2z$	$=$ $=$	$12$ $12$
$+$ $+$ $3 x - y + 2 z$ $3x-y+2z$	$=$ $=$	$5$ $5$	$+ 2 z - 2 z$ $+2z-2z$ cancels out

$11 x + 3 y$ $11x+3y$	$=$ $=$	$17$ $17$	Equation $B$ $B$

Next, multiply the values of equation

A

$A$ by

3

$3$ and add the product to equation

B

$B$ to find the value of $x$ $x$ .

$3 \times$ $3xx$ $(9 x - y)$ $(9x-y)$	$=$ $=$	$3 \times$ $3xx$ $7$ $7$	Multiply equation $A$ $A$ by $3$ $3$
$27 x - 3 y$ $27x-3y$	$=$ $=$	$21$ $21$

$27 x - 3 y$ $27x-3y$	$=$ $=$	$21$ $21$
$+$ $+$ $11 x + 3 y$ $11x+3y$	$=$ $=$	$17$ $17$

$38 x$ $38x$	$=$ $=$	$38$ $38$	$- 3 y + 3 y$ $-3y+3y$ cancels out
$38 x$ $38x$ $\div 38$ $div38$	$=$ $=$	$38$ $38$ $\div 38$ $div38$	Divide both sides by $38$ $38$
$x$ $x$	$=$ $=$	$1$ $1$

Now, substitute the value of $x$ $x$ into equation

A

$A$ to solve for $y$ $y$ .

$9$ $9$ $x$ $x$ $- y$ $-y$	$=$ $=$	$7$ $7$	Equation $A$ $A$
$9 ($ $9($ $1$ $1$ $) - y$ $)-y$	$=$ $=$	$7$	$x=1$
$9-y$	$=$	$7$
$9-y$ $-9$	$=$	$7$ $-9$	Subtract $9$ from both sides
$-y$	$=$	$-2$
$-y$ $times(-1)$	$=$	$-2$ $times(-1)$	Multiply both sides by $-1$
$y$	$=$	$2$

Finally, substitute the value of $x$ and $y$ into equation

$1$ to solve for $z$ .

$3$ $x$ $-$ $y$ $+2z$	$=$	$5$	Equation $1$
$3($ $1$ $)-$ $2$ $+2z$	$=$	$5$	$x=1$ and $y=2$
$3-2+2z$	$=$	$5$
$1+2z$	$=$	$5$
$1+2z$ $-1$	$=$	$5$ $-1$	Subtract $1$ from both sides
$2z$	$=$	$4$
$2z$ $div2$	$=$	$4$ $div2$	Divide both sides by $2$
$z$	$=$	$2$

$x=1$

$y =2$

$z =2$

Question 2 of 5

Solve the following systems of equations by elimination.

$3x+5y+z=5$

$9x+2y+7z=32$

$6x+3y+4z=19$

$x=$ (3)

$y=$ (-1)

$z=$ (1)

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Elimination Method

$1)$ make sure 2 equations have a variable with same coefficients but opposite signs
$2)$ add or subtract the equations so that one variable is cancelled
$3)$ solve for the variable that remains
$4)$ substitute known value to one of the equations to solve for the other variable

First, label the equations

$1$ ,

$2$ and

$3$ respectively.

$3x+5y+z$	$=$	$5$	Equation $1$
$9x+2y+7z$	$=$	$32$	Equation $2$
$6x+3y+4z$	$=$	$19$	Equation $3$

Next, multiply the values of equation

$1$ by

$3$ and subtract equation

$1$ from the product. Label the result as equation

$A$ .

$3xx$ $(3x+5y+z)$	$=$	$3xx$ $5$	Multiply equation $1$ by $3$
$9x+15y+3z$	$=$	$15$

$9x+15y+3z$	$=$	$15$
$-$ $9x-2y-7z$	$=$	$32$	$9x-9x$ cancels out

$13y-4z$	$=$	$-17$	Equation $A$

Then, multiply the values of equation

$1$ by

$2$ and subtract equation

$3$ from the product. Label the result as equation

$B$ .

$2xx$ $(3x+5y+z)$	$=$	$2xx$ $5$	Multiply equation $1$ by $2$
$6x+10y+2z$	$=$	$10$

$6x+10y+2z$	$=$	$10$
$-$ $6x-3y-4z$	$=$	$19$	$6x-6x$ cancels out

$7y-2z$	$=$	$-9$	Equation $B$

Now, multiply the values of equation

$B$ by

$2$ and subtract equation

$A$ from the product to find the value of $y$ .

$2xx$ $(7y-2z)$	$=$	$2xx$ $-9$	Multiply equation $B$ by $2$
$14y-4z$	$=$	$-18$

$14y-4z$	$=$	$-18$
$-$ $13y+4z$	$=$	$17$	$-4z+4z$ cancels out

$y$	$=$	$-1$

Now, substitute the value of $y$ into equation

$B$ to solve for $z$ .

$7$ $y$ $-2z$	$=$	$-9$	Equation $B$
$7($ $-1$ $)-2z$	$=$	$-9$	$y=-1$
$-7-2z$	$=$	$-9$
$-7-2z$ $+7$	$=$	$-9$ $+7$	Add $7$ to both sides
$-2z$	$=$	$-2$
$-2z$ $div(-2)$	$=$	$-2$ $div(-2)$	Divide both sides by $-2$
$z$	$=$	$1$

Finally, substitute the value of $y$ and $z$ into equation

$1$ to solve for $x$ .

$3x+5$ $y$ $+$ $z$	$=$	$5$	Equation $1$
$3x+5($ $-1$ $)+$ $1$	$=$	$5$	$y=-1$ and $z=1$
$3x-5+1$	$=$	$5$
$3x-4$	$=$	$5$
$3x-4$ $+4$	$=$	$5$ $+4$	Add $4$ to both sides
$3x$	$=$	$9$
$3x$ $div3$	$=$	$9$ $div3$	Divide both sides by $3$
$x$	$=$	$3$

$x=3$

$y =-1$

$z =1$

Question 3 of 5

Solve the following systems of equations by elimination.

$x+2y+5z=2$

$3x+4y-4z=21$

$2x-y+z=3$

$x=$ (3)

$y=$ (2)

$z=$ (-1)

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Elimination Method

$1)$ make sure 2 equations have a variable with same coefficients but opposite signs
$2)$ add or subtract the equations so that one variable is cancelled
$3)$ solve for the variable that remains
$4)$ substitute known value to one of the equations to solve for the other variable

First, label the equations

$1$ ,

$2$ and

$3$ respectively.

$x+2y+5z$	$=$	$2$	Equation $1$
$3x+4y-4z$	$=$	$21$	Equation $2$
$2x-y+z$	$=$	$3$	Equation $3$

Next, multiply the values of equation

$1$ by

$3$ and subtract equation

$2$ from the product. Label the result as equation

$A$ .

$3xx$ $(x+2y+5z)$	$=$	$3xx$ $2$	Multiply equation $1$ by $3$
$3x+6y+15z$	$=$	$6$

$3x+6y+15z$	$=$	$6$
$-$ $3x-4y+4z$	$=$	$21$	$3x-3x$ cancels out

$2y+19z$	$=$	$-15$	Equation $A$

Then, multiply the values of equation

$1$ by

$2$ and subtract equation

$3$ from the product. Label the result as equation

$B$ .

$2xx$ $(x+2y+5z)$	$=$	$2xx$ $2$	Multiply equation $1$ by $2$
$2x+4y+10z$	$=$	$4$

$2x+4y+10z$	$=$	$4$
$-$ $2x+y-z$	$=$	$-3$	$2x-2x$ cancels out

$5y+9z$	$=$	$1$	Equation $B$

Now, multiply the values of equation

$A$ by

$5$ and the values of equation

$B$ by

$2$ , then get their difference to find the value of $z$ .

$5xx$ $(2y+19z)$	$=$	$5xx$ $-15$	Multiply equation $A$ by $5$
$10y+95z$	$=$	$-75$

$2xx$ $(5y+9z)$	$=$	$2xx$ $1$	Multiply equation $B$ by $2$
$10y+18z$	$=$	$2$

$10y+18z$	$=$	$2$
$-$ $10y-95z$	$=$	$75$	$10y-10y$ cancels out

$-77z$	$=$	$77$
$-77z$ $div(-77)$	$=$	$77$ $div(-77)$	Divide both sides by $-77$
$z$	$=$	$-1$

Now, substitute the value of $z$ into equation

$B$ to solve for $y$ .

$5y+9$ $z$	$=$	$1$	Equation $B$
$5y+9($ $-1$ $)$	$=$	$1$	$z=-1$
$5y-9$	$=$	$1$
$5y-9$ $+9$	$=$	$1$ $+9$	Add $9$ to both sides
$5y$	$=$	$10$
$5y$ $div5$	$=$	$10$ $div5$	Divide both sides by $5$
$y$	$=$	$2$

Finally, substitute the value of $y$ and $z$ into equation

$1$ to solve for $x$ .

$x+2$ $y$ $+5$ $z$	$=$	$2$	Equation $1$
$x+2($ $2$ $)+5($ $-1$ $)$	$=$	$2$	$y=2$ and $z=-1$
$x+4-5$	$=$	$2$
$x-1$	$=$	$2$
$x-1$ $+1$	$=$	$2$ $+1$	Add $1$ to both sides
$x$	$=$	$3$

$x=3$

$y =2$

$z =-1$

Question 4 of 5

Solve the following systems of equations by elimination.

$-2x-y+5z=5$

$3x+2y-z=7$

$-4x+4y+2z=12$

$x=$ (1)

$y=$ (3)

$z=$ (2)

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Elimination Method

$1)$ make sure 2 equations have a variable with same coefficients but opposite signs
$2)$ add or subtract the equations so that one variable is cancelled
$3)$ solve for the variable that remains
$4)$ substitute known value to one of the equations to solve for the other variable

First, label the equations

$1$ ,

$2$ and

$3$ respectively.

$-2x-y+5z$	$=$	$5$	Equation $1$
$3x+2y-z$	$=$	$7$	Equation $2$
$-4x+4y+2z$	$=$	$12$	Equation $3$

Next, multiply the values of equation

$1$ by

$2$ and add equation

$2$ to the product. Label the result as equation

$A$ .

$2xx$ $(-2x-y+5z)$	$=$	$2xx$ $5$	Multiply equation $1$ by $2$
$-4x-2y+10z$	$=$	$10$

$-4x-2y+10z$	$=$	$10$
$+$ $3x+2y-z$	$=$	$7$	$-2y+2y$ cancels out

$-x+9z$	$=$	$17$	Equation $A$

Then, multiply the values of equation

$2$ by

$2$ and subtract equation

$3$ from the product. Label the result as equation

$B$ .

$2xx$ $(3x+2y-z)$	$=$	$2xx$ $7$	Multiply equation $2$ by $2$
$6x+4y-2z$	$=$	$14$

$6x+4y-2z$	$=$	$14$
$-$ $4x-4y-2z$	$=$	$-12$	$4y-4y$ cancels out

$10x-4z$	$=$	$2$	Equation $B$

Now, multiply the values of equation

$A$ by

$10$ and add the product to equation

$A$ to find the value of $z$ .

$10xx$ $(-x+9z)$	$=$	$10xx$ $17$	Multiply equation $A$ by $10$
$-10x+90z$	$=$	$170$

$10x-4z$	$=$	$2$
$+$ $-10x+90z$	$=$	$170$	$10x-10x$ cancels out

$86z$	$=$	$172$
$86z$ $div86$	$=$	$172$ $div86$	Divide both sides by $86$
$z$	$=$	$2$

Now, substitute the value of $z$ into equation

$B$ to solve for $x$ .

$10x-4$ $z$	$=$	$2$	Equation $B$
$10x-4($ $2$ $)$	$=$	$2$	$y=-1$
$10x-8$	$=$	$2$
$10x-8$ $+8$	$=$	$2$ $+8$	Add $8$ to both sides
$10x$	$=$	$10$
$10x$ $div10$	$=$	$10$ $div10$	Divide both sides by $10$
$x$	$=$	$1$

Finally, substitute the value of $x$ and $z$ into equation

$2$ to solve for $y$ .

$3$ $x$ $+2y-$ $z$	$=$	$7$	Equation $2$
$3($ $1$ $)+2y-$ $2$	$=$	$7$	$x=1$ and $z=2$
$3+2y-2$	$=$	$7$
$1+2y$	$=$	$7$
$1+2y$ $-1$	$=$	$7$ $-1$	Subtract $1$ from both sides
$2y$	$=$	$6$
$2y$ $div2$	$=$	$6$ $div2$	Divide both sides by $2$
$y$	$=$	$3$

$x=1$

$y =3$

$z =2$

Question 5 of 5

Solve the following systems of equations by elimination.

$2a-b-4c=1$

$a+6b-c=8$

$3a+4b-2c=11$

$a=$ (3)

$b=$ (1)

$c=$ (1)

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Elimination Method

$1)$ make sure 2 equations have a variable with same coefficients but opposite signs
$2)$ add or subtract the equations so that one variable is cancelled
$3)$ solve for the variable that remains
$4)$ substitute known value to one of the equations to solve for the other variable

First, label the equations

$1$ ,

$2$ and

$3$ respectively.

$2a-b-4c$	$=$	$1$	Equation $1$
$a+6b-c$	$=$	$8$	Equation $2$
$3a+4b-2c$	$=$	$11$	Equation $3$

Next, multiply the values of equation

$2$ by

$2$ and subtract equation

$1$ from the product. Label the result as equation

$A$ .

$2xx$ $(a+6b-c)$	$=$	$2xx$ $8$	Multiply equation $2$ by $2$
$2a+12b-2c$	$=$	$16$

$2a+12b-2c$	$=$	$16$
$-$ $2a+b+4c$	$=$	$1$	$2a-2a$ cancels out

$13b+2c$	$=$	$15$	Equation $A$

Then, multiply the values of equation

$2$ by

$3$ and subtract equation

$3$ from the product. Label the result as equation

$B$ .

$3xx$ $(a+6b-c)$	$=$	$3xx$ $8$	Multiply equation $2$ by $3$
$3a+18b-3c$	$=$	$24$

$3a+18b-3c$	$=$	$24$
$-$ $3a-4b+2c$	$=$	$-11$	$6x-6x$ cancels out

$14b-c$	$=$	$13$	Equation $B$

Now, multiply the values of equation

$B$ by

$2$ and add equation

$A$ to the product to find the value of $y$ .

$2xx$ $(14b-c)$	$=$	$2xx$ $13$	Multiply equation $B$ by $2$
$28b-2c$	$=$	$26$

$28b-2c$	$=$	$26$
$+$ $13b+2c$	$=$	$15$	$-4z+4z$ cancels out

$41b$	$=$	$41$
$41b$ $div41$	$=$	$41$ $div41$	Divide both sides by $41$
$b$	$=$	$1$

Now, substitute the value of $b$ into equation

$A$ to solve for $c$ .

$13$ $b$ $+2c$	$=$	$15$	Equation $A$
$13($ $1$ $)+2c$	$=$	$15$	$y=1$
$13+2c$	$=$	$15$
$13+2c$ $-13$	$=$	$15$ $-13$	Subtract $13$ from both sides
$2c$	$=$	$2$
$2c$ $div2$	$=$	$2$ $div2$	Divide both sides by $2$
$c$	$=$	$1$

Finally, substitute the value of $b$ and $c$ into equation

$2$ to solve for $a$ .

$a+6$ $b$ $-$ $c$	$=$	$8$	Equation $1$
$a+6($ $1$ $)-$ $1$	$=$	$8$	$b=1$ and $c=1$
$a+6-1$	$=$	$8$
$a+5$	$=$	$8$
$a+5$ $-5$	$=$	$8$ $-5$	Subtract $5$ from both sides
$a$	$=$	$3$

$a=3$

$b =1$

$c =1$

3 Variable Systems of Equations – Elimination Method

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