3 Variable Simultaneous Equations – Elimination Method

Question 1 of 5

Solve the following simultaneous 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 $A$ $A$
$9 ($ $9($ $1$ $1$ $) - y$ $)-y$	$=$ $=$	$7$ $7$	$x = 1$ $x=1$
$9 - y$ $9-y$	$=$ $=$	$7$ $7$
$9 - y$ $9-y$ $- 9$ $-9$	$=$ $=$	$7$ $7$ $- 9$ $-9$	Subtract $9$ $9$ from both sides
$- y$ $-y$	$=$ $=$	$- 2$ $-2$
$- y$ $-y$ $\times (- 1)$ $times(-1)$	$=$ $=$	$- 2$ $-2$ $\times (- 1)$ $times(-1)$	Multiply both sides by $- 1$ $-1$
$y$ $y$	$=$ $=$	$2$ $2$

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

1

$1$ to solve for $z$ $z$ .

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

x = 1

$x=1$

y = 2

$y =2$

z = 2

$z =2$

Question 2 of 5

Solve the following simultaneous equations by elimination.

3 x + 5 y + z = 5

$3x+5y+z=5$

9 x + 2 y + 7 z = 32

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

6 x + 3 y + 4 z = 19

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

$x =$ $x=$ (3)

$y =$ $y=$ (-1)

$z =$ $z=$ (1)

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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.

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

Next, multiply the values of equation

1

$1$ by

3

$3$ and subtract equation

1

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

A

$A$ .

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

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

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

Then, multiply the values of equation

1

$1$ by

2

$2$ and subtract equation

3

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

B

$B$ .

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

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

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

Now, multiply the values of equation

B

$B$ by

2

$2$ and subtract equation

A

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

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

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

$y$ $y$	$=$ $=$	$- 1$ $-1$

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

B

$B$ to solve for $z$ $z$ .

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

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

1

$1$ to solve for $x$ $x$ .

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

x = 3

$x=3$

y = - 1

$y =-1$

z = 1

$z =1$

Question 3 of 5

Solve the following simultaneous equations by elimination.

x + 2 y + 5 z = 2

$x+2y+5z=2$

3 x + 4 y - 4 z = 21

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

2 x - y + z = 3

$2x-y+z=3$

$x =$ $x=$ (3)

$y =$ $y=$ (2)

$z =$ $z=$ (-1)

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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.

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

Next, multiply the values of equation

1

$1$ by

3

$3$ and subtract equation

2

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

A

$A$ .

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

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

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

Then, multiply the values of equation

1

$1$ by

2

$2$ and subtract equation

3

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

B

$B$ .

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

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

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

Now, multiply the values of equation

A

$A$ by

5

$5$ and the values of equation

B

$B$ by

2

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

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

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

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

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

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

B

$B$ to solve for $y$ $y$ .

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

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

1

$1$ to solve for $x$ $x$ .

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

x = 3

$x=3$

y = 2

$y =2$

z = - 1

$z =-1$

Question 4 of 5

Solve the following simultaneous equations by elimination.

- 2 x - y + 5 z = 5

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

3 x + 2 y - z = 7

$3x+2y-z=7$

- 4 x + 4 y + 2 z = 12

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

$x =$ $x=$ (1)

$y =$ $y=$ (3)

$z =$ $z=$ (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.

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

Next, multiply the values of equation

1

$1$ by

2

$2$ and add equation

2

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

A

$A$ .

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

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

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

Then, multiply the values of equation

2

$2$ by

2

$2$ and subtract equation

3

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

B

$B$ .

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

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

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

Now, multiply the values of equation

A

$A$ by

10

$10$ and add the product to equation

A

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

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

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

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

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

B

$B$ to solve for $x$ $x$ .

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

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

2

$2$ to solve for $y$ $y$ .

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

x = 1

$x=1$

y = 3

$y =3$

z = 2

$z =2$

Question 5 of 5

Solve the following simultaneous equations by elimination.

2 a - b - 4 c = 1

$2a-b-4c=1$

a + 6 b - c = 8

$a+6b-c=8$

3 a + 4 b - 2 c = 11

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

$a =$ $a=$ (3)

$b =$ $b=$ (1)

$c =$ $c=$ (1)

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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.

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

Next, multiply the values of equation

2

$2$ by

2

$2$ and subtract equation

1

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

A

$A$ .

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

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

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

Then, multiply the values of equation

2

$2$ by

3

$3$ and subtract equation

3

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

B

$B$ .

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

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

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

Now, multiply the values of equation

B

$B$ by

2

$2$ and add equation

A

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

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

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

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

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

A

$A$ to solve for $c$ $c$ .

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

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

2

$2$ to solve for $a$ $a$ .

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

a = 3

$a=3$

b = 1

$b =1$

c = 1

$c =1$

3 Variable Simultaneous Equations – Elimination Method

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Elimination Method

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Elimination Method

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