3 Variable Systems of Equations – Substitution Method

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Question 1 of 2

Solve the following systems of equations by substitution.

a - 2 b + c = - 4

$a-2b+c=-4$

2 a + 2 b - c = 10

$2a+2b-c=10$

- 3 a - b - 4 c = 9

$-3a-b-4c=9$

$a =$ $a=$ (2)

$b =$ $b=$ (1)

$c =$ $c=$ (-4)

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Substitution Method

$1)$ $1)$ Make one variable the subject in one equation
$2)$ $2)$ Substitute this variable in to the other two equations
$3)$ $3)$ Solve these two equations using substitution and find $2$ $2$ variables
$4)$ $4)$ Substitute the $2$ $2$ variables into the original equation

First, label the equations

1

$1$ ,

2

$2$ and

3

$3$ respectively.

$a - 2 b + c$ $a-2b+c$	$=$ $=$	$- 4$ $-4$	Equation $1$ $1$
$2 a + 2 b - c$ $2a+2b-c$	$=$ $=$	$10$ $10$	Equation $2$ $2$
$- 3 a - b - 4 c$ $-3a-b-4c$	$=$ $=$	$9$ $9$	Equation $3$ $3$

Next, make $a$ $a$ the subject in Equation

1

$1$ .

$a - 2 b + c$ $a-2b+c$	$=$ $=$	$- 4$ $-4$
$a - 2 b + c$ $a-2b+c$ $+ (2 b - c)$ $+(2b-c)$	$=$ $=$	$- 4$ $-4$ $+ (2 b - c)$ $+(2b-c)$	Add $2 b - c$ $2b-c$ to both sides
$a$ $a$	$=$ $=$	$2 b - c - 4$ $2b-c-4$	Simplify

Substitute Equation

1

$1$ ’s $a$ $a$ into both Equations

2

$2$ and

3

$3$ . Label the results as Equations

A

$A$ and

B

$B$

$2$ $2$ $a$ $a$ $+ 2 b - c$ $+2b-c$	$=$ $=$	$10$ $10$	Equation $2$ $2$
$2 ($ $2($ $2 b - c - 4$ $2b-c-4$ $) + 2 b - c$ $)+2b-c$	$=$ $=$	$10$ $10$	$a = 2 b - c - 4$ $a=2b-c-4$
$4 b - 2 c - 8 + 2 b - c$ $4b-2c-8+2b-c$	$=$ $=$	$10$ $10$	Distribute $2$ $2$ inside the parenthesis
$6 b - 3 c - 8$ $6b-3c-8$	$=$ $=$	$10$ $10$	Simplify
$6 b - 3 c - 8$ $6b-3c-8$ $+ 8$ $+8$	$=$ $=$	$10$ $10$ $+ 8$ $+8$	Add $8$ $8$ to both sides
$6 b - 3 c$ $6b-3c$	$=$ $=$	$18$ $18$	Equation $A$ $A$

$- 3$ $-3$ $a$ $a$ $- b - 4 c$ $-b-4c$	$=$ $=$	$9$ $9$	Equation $3$ $3$
$- 3 ($ $-3($ $2 b - c - 4$ $2b-c-4$ $) - b - 4 c$ $)-b-4c$	$=$ $=$	$9$ $9$	$a = 2 b - c - 4$ $a=2b-c-4$
$- 6 b + 3 c + 12 - b - 4 c$ $-6b+3c+12-b-4c$	$=$ $=$	$9$ $9$	Distribute $2$ $2$ inside the parenthesis
$- 7 b - c + 12$ $-7b-c+12$	$=$ $=$	$9$ $9$	Simplify
$- 7 b - c + 12$ $-7b-c+12$ $- 12$ $-12$	$=$ $=$	$9$ $9$ $- 12$ $-12$	Subtract $12$ $12$ from both sides
$- 7 b - c$ $-7b-c$	$=$ $=$	$- 3$ $-3$	Equation $B$ $B$

Next, make $c$ $c$ the subject in Equation

B

$B$ .

$- 7 b - c$ $-7b-c$	$=$ $=$	$- 3$ $-3$	Equation $B$ $B$
$- 7 b - c$ $-7b-c$ $+ c$ $+c$	$=$ $=$	$- 3$ $-3$ $+ c$ $+c$	Add $c$ $c$ to both sides
$- 7 b$ $-7b$	$=$ $=$	$- 3 + c$ $-3+c$
$- 7 b$ $-7b$ $+ 3$ $+3$	$=$ $=$	$- 3 + c$ $-3+c$ $+ 3$ $+3$	Add $3$ $3$ to both sides
$- 7 b + 3$ $-7b+3$	$=$ $=$	$c$ $c$
$c$ $c$	$=$ $=$	$- 7 b + 3$ $-7b+3$

Now, substitute $c = - 7 b + 3$ $c=-7b+3$ into Equation

A

$A$ and solve for $b$ $b$ .

$6 b - 3$ $6b-3$ $c$ $c$	$=$ $=$	$18$ $18$	Equation $A$ $A$
$6 b - 3 ($ $6b-3($ $- 7 b + 3$ $-7b+3$ $)$ $)$	$=$ $=$	$18$ $18$	$c = - 7 b + 3$ $c=-7b+3$
$6 b + 21 b - 9$ $6b+21b-9$	$=$ $=$	$18$ $18$	Distribute $3$ $3$ inside the parenthesis
$27 b - 9$ $27b-9$	$=$ $=$	$18$ $18$	Simplify
$27 b - 9$ $27b-9$ $+ 9$ $+9$	$=$ $=$	$18$ $18$ $+ 9$ $+9$	Add $9$ $9$ to both sides
$27 b$ $27b$	$=$ $=$	$27$ $27$
$27 b$ $27b$ $\div 27$ $div27$	$=$ $=$	$27$ $27$ $\div 27$ $div27$	Divide both sides by $27$ $27$
$b$ $b$	$=$ $=$	$1$ $1$

Next, substitute the value of $b$ $b$ into Equation

B

$B$ and solve for $c$ $c$

$c$ $c$	$=$ $=$	$- 7$ $-7$ $b$ $b$ $+ 3$ $+3$	Equation $B$ $B$
$c$ $c$	$=$ $=$	$- 7 ($ $-7($ $1$ $1$ $) + 3$ $)+3$	$b = 1$ $b=1$
$c$ $c$	$=$ $=$	$- 7 + 3$ $-7+3$
$c$ $c$	$=$ $=$	$- 4$ $-4$

Finally, substitute the known values of $b$ $b$ and $c$ $c$ into Equation

1

$1$ to solve for $a$ $a$ .

$a - 2$ $a-2$ $b$ $b$ $+$ $+$ $c$ $c$	$=$ $=$	$- 4$ $-4$	Equation $1$ $1$
$a - 2 ($ $a-2($ $1$ $1$ $) +$ $)+$ $- 4$ $-4$	$=$ $=$	$- 4$ $-4$	$b = 1$ $b=1$ and $c = - 4$ $c=-4$
$a - 2 - 4$ $a-2-4$	$=$ $=$	$- 4$ $-4$
$a - 6$ $a-6$	$=$ $=$	$- 4$ $-4$	Simplify
$a - 6$ $a-6$ $+ 6$ $+6$	$=$ $=$	$- 4$ $-4$ $+ 6$ $+6$	Add $6$ $6$ to both sides
$a$ $a$	$=$ $=$	$2$ $2$

a = 2

$a=2$

b = 1

$b =1$

c = - 4

$c =-4$

Question 2 of 2

Solve the following systems of equations by substitution.

3 x - 3 y + z = - 2

$3x-3y+z=-2$

2 y - z = - 5

$2y-z=-5$

x + 4 y + 2 z = - 9

$x+4y+2z=-9$

$x =$ $x=$ (-3)

$y =$ $y=$ (-2)

$z =$ $z=$ (1)

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Substitution Method

$1)$ $1)$ Make one variable the subject in one equation
$2)$ $2)$ Substitute this variable in to the other two equations
$3)$ $3)$ Solve these two equations using substitution and find $2$ $2$ variables
$4)$ $4)$ Substitute the $2$ $2$ variables into the original equation

First, label the equations

1

$1$ ,

2

$2$ and

3

$3$ respectively.

$3 x - 3 y + z$ $3x-3y+z$	$=$ $=$	$- 2$ $-2$	Equation $1$ $1$
$2 y - z$ $2y-z$	$=$ $=$	$- 5$ $-5$	Equation $2$ $2$
$x + 4 y + 2 z$ $x+4y+2z$	$=$ $=$	$- 9$ $-9$	Equation $3$ $3$

Next, make $z$ $z$ the subject in Equation

1

$1$ .

$2 y - z$ $2y-z$	$=$ $=$	$- 5$ $-5$
$2 y - z$ $2y-z$ $+ 5$ $+5$	$=$ $=$	$- 5$ $-5$ $+ 5$ $+5$	Add $5$ $5$ to both sides
$2 y - z + 5$ $2y-z+5$	$=$ $=$	$0$ $0$
$2 y + 5 - z$ $2y+5-z$ $+ z$ $+z$	$=$ $=$	$0$ $0$ $+ z$ $+z$	Add $z$ $z$ to both sides
$2 y + 5$ $2y+5$	$=$ $=$	$z$ $z$
$z$ $z$	$=$ $=$	$2 y + 5$ $2y+5$	Simplify

Substitute $z = 2 y + 5$ $z=2y+5$ into both Equations

1

$1$ and

3

$3$ . Label the results as Equations

A

$A$ and

B

$B$

$3 x - 3 y +$ $3x-3y+$ $z$ $z$	$=$ $=$	$- 2$ $-2$	Equation $1$ $1$
$3 x - 3 y +$ $3x-3y+$ $2 y + 5$ $2y+5$	$=$ $=$	$- 2$ $-2$	$z = 2 y + 5$ $z=2y+5$
$3 x - y + 5$ $3x-y+5$	$=$ $=$	$- 2$ $-2$	Combine like terms
$3 x - y + 5$ $3x-y+5$ $- 5$ $-5$	$=$ $=$	$- 2$ $-2$ $- 5$ $-5$	Subtract $5$ $5$ from both sides
$3 x - y$ $3x-y$	$=$ $=$	$- 7$ $-7$	Equation $A$ $A$

$x + 4 y + 2$ $x+4y+2$ $z$ $z$	$=$ $=$	$- 9$ $-9$	Equation $3$ $3$
$x + 4 y + 2 ($ $x+4y+2($ $2 y + 5$ $2y+5$ $)$ $)$	$=$ $=$	$- 9$ $-9$	$z = 2 y + 5$ $z=2y+5$
$x + 4 y + 4 y + 10$ $x+4y+4y+10$	$=$ $=$	$- 9$ $-9$	Distribute $2$ $2$ inside the parenthesis
$x + 8 y + 10$ $x+8y+10$	$=$ $=$	$- 9$ $-9$	Combine like terms
$x + 8 y + 10$ $x+8y+10$ $- 10$ $-10$	$=$ $=$	$- 9$ $-9$ $- 10$ $-10$	Subtract $10$ $10$ from both sides
$x + 8 y$ $x+8y$	$=$ $=$	$- 19$ $-19$	Equation $B$ $B$

Next, make $y$ $y$ the subject in Equation

A

$A$ .

$3 x - y$ $3x-y$	$=$ $=$	$- 7$ $-7$	Equation $A$ $A$
$3 x - y$ $3x-y$ $+ 7$ $+7$	$=$ $=$	$- 7$ $-7$ $+ 7$ $+7$	Add $7$ $7$ to both sides
$3 x + 7 - y$ $3x+7-y$	$=$ $=$	$0$ $0$
$3 x + 7 - y$ $3x+7-y$ $+ y$ $+y$	$=$ $=$	$0$ $0$ $+ y$ $+y$	Add $y$ $y$ to both sides
$3 x + 7$ $3x+7$	$=$ $=$	$y$ $y$
$y$ $y$	$=$ $=$	$3 x + 7$ $3x+7$

Now, substitute $y = 3 x + 7$ $y=3x+7$ into Equation

B

$B$ and solve for $x$ $x$ .

$x + 8$ $x+8$ $y$ $y$	$=$ $=$	$- 19$ $-19$	Equation $B$ $B$
$x + 8 ($ $x+8($ $3 x + 7$ $3x+7$ $)$ $)$	$=$ $=$	$- 19$ $-19$	$y = 3 x + 7$ $y=3x+7$
$x + 24 x + 56$ $x+24x+56$	$=$ $=$	$- 19$ $-19$	Distribute $8$ $8$ inside the parenthesis
$25 x + 56$ $25x+56$	$=$ $=$	$- 19$ $-19$	Combine like terms
$25 x + 56$ $25x+56$ $- 56$ $-56$	$=$ $=$	$- 19$ $-19$ $- 56$ $-56$	Subtract $56$ $56$ from both sides
$25 x$ $25x$	$=$ $=$	$- 75$ $-75$
$27 x$ $27x$ $\div 25$ $div25$	$=$ $=$	$- 75$ $-75$ $\div 25$ $div25$	Divide both sides by $25$ $25$
$x$ $x$	$=$ $=$	$- 3$ $-3$

Next, substitute the value of $x$ $x$ into Equation

A

$A$ and solve for $y$ $y$

$3$ $3$ $x$ $x$ $- y$ $-y$	$=$ $=$	$- 7$ $-7$	Equation $A$ $A$
$3 ($ $3($ $- 3$ $-3$ $) - y$ $)-y$	$=$ $=$	$- 7$ $-7$	$x = - 3$ $x=-3$
$- 9 - y$ $-9-y$	$=$ $=$	$- 7$ $-7$
$- 9 - y$ $-9-y$ $+ 9$ $+9$	$=$ $=$	$- 7$ $-7$ $+ 9$ $+9$	Add $9$ $9$ to 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 known value of $y$ $y$ into Equation

2

$2$ to solve for $z$ $z$ .

$2$ $2$ $y$ $y$ $- z$ $-z$	$=$ $=$	$- 5$ $-5$	Equation $2$ $2$
$2 ($ $2($ $- 2$ $-2$ $) - z$ $)-z$	$=$ $=$	$- 5$ $-5$	$y = - 2$ $y=-2$
$- 4 - z$ $-4-z$	$=$ $=$	$- 5$ $-5$
$- 4 - z$ $-4-z$ $+ 4$ $+4$	$=$ $=$	$- 5$ $-5$ $+ 4$ $+4$	Add $4$ $4$ to both sides
$- z$ $-z$	$=$ $=$	$- 1$ $-1$
$- z$ $-z$ $\times (- 1)$ $times(-1)$	$=$ $=$	$- 1$ $-1$ $\times (- 1)$ $times(-1)$	Multiply both sides by $- 1$ $-1$
$z$ $z$	$=$ $=$	$1$ $1$

x = - 3

$x=-3$

y = - 2

$y =-2$

z = 1

$z =1$

3 Variable Systems of Equations – Substitution Method

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1. Question

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Fantastic!

Substitution Method

2. Question

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Correct!

Substitution Method

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