3 Variable Systems of Equations – Substitution Method

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Question 1 of 2

Solve the following systems of equations by substitution.

$a-2b+c=-4$

$2a+2b-c=10$

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

$a=$ (2)

$b=$ (1)

$c=$ (-4)

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Substitution Method

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

First, label the equations

$1$ ,

$2$ and

$3$ respectively.

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

Next, make $a$ the subject in Equation

$1$ .

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

Substitute Equation

$1$ ’s $a$ into both Equations

$2$ and

$3$ . Label the results as Equations

$A$ and

$B$

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

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

Next, make $c$ the subject in Equation

$B$ .

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

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

$A$ and solve for $b$ .

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

Next, substitute the value of $b$ into Equation

$B$ and solve for $c$

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

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

$1$ to solve for $a$ .

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

$a=2$

$b =1$

$c =-4$

Question 2 of 2

Solve the following systems of equations by substitution.

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

$2y-z=-5$

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

$x=$ (-3)

$y=$ (-2)

$z=$ (1)

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Substitution Method

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

First, label the equations

$1$ ,

$2$ and

$3$ respectively.

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

Next, make $z$ the subject in Equation

$1$ .

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

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

$1$ and

$3$ . Label the results as Equations

$A$ and

$B$

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

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

Next, make $y$ the subject in Equation

$A$ .

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

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

$B$ and solve for $x$ .

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

Next, substitute the value of $x$ into Equation

$A$ and solve for $y$

$3$ $x$ $-y$	$=$	$-7$	Equation $A$
$3($ $-3$ $)-y$	$=$	$-7$	$x=-3$
$-9-y$	$=$	$-7$
$-9-y$ $+9$	$=$	$-7$ $+9$	Add $9$ to both sides
$-y$	$=$	$2$
$-y$ $times(-1)$	$=$	$2$ $times(-1)$	Multiply both sides by $-1$
$y$	$=$	$-2$

Finally, substitute the known value of $y$ into Equation

$2$ to solve for $z$ .

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

$x=-3$

$y =-2$

$z =1$

3 Variable Systems of Equations – Substitution Method

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Substitution Method

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Substitution Method

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