Substitution Method 4

Question 1 of 4

Solve the following systems of equations by substitution.

b = 2 a + 2

$b=2a+2$

2 a + 5 b = 10

$2a+5b=10$

$a =$ $a=$ (0)

$b =$ $b=$ (2)

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Substitution Method

$1)$ $1)$ make one variable the subject
$2)$ $2)$ substitute into second equation
$3)$ $3)$ solve the second equation
$4)$ $4)$ substitute back

First, label the two equations

1

$1$ and

2

$2$ respectively.

$b$ $b$	$=$ $=$	$2 a + 2$ $2a+2$	Equation $1$ $1$
$2 a + 5 b$ $2a+5b$	$=$ $=$	$10$ $10$	Equation $2$ $2$

Next, substitute $b$ $b$ from Equation

1

$1$ into Equation

2

$2$ .

$2 a + 5$ $2a+5$ $b$ $b$	$=$ $=$	$10$ $10$	Equation $2$ $2$
$2 a + 5$ $2a+5$ $(2 a + 2)$ $(2a+2)$	$=$ $=$	$10$ $10$	$b = 2 a + 2$ $b=2a+2$
$2 a + 10 a + 10$ $2a+10a+10$	$=$ $=$	$10$ $10$	Distribute $5$ $5$ inside the parenthesis
$12 a + 10$ $12a+10$	$=$ $=$	$10$ $10$	Simplify
$12 a + 10$ $12a+10$ $- 10$ $-10$	$=$ $=$	$10$ $10$ $- 10$ $-10$	Solve for $a$ $a$
$12 a$ $12a$	$=$ $=$	$0$ $0$
$12 a$ $12a$ $\div 12$ $div12$	$=$ $=$	$0$ $0$ $\div 12$ $div12$	Divide both sides by $12$ $12$
$a$ $a$	$=$ $=$	$0$ $0$

Now, substitute the value of $a$ $a$ into Equation

1

$1$

$b$ $b$	$=$ $=$	$2$ $2$ $a$ $a$ $+ 2$ $+2$	Equation $1$ $1$
$b$ $b$	$=$ $=$	$2$ $2$ $(0)$ $(0)$ $+ 2$ $+2$	$a = 0$ $a=0$
$b$ $b$	$=$ $=$	$0 + 2$ $0+2$
$b$ $b$	$=$ $=$	$2$ $2$

a = 0, b = 2

$a=0, b=2$

Question 2 of 4

Solve the following systems of equations by substitution.

2 m + 5 n = 5

$2m+5n=5$

m - 5 n = 10

$m-5n=10$

$m =$ $m=$ (5)

$n =$ $n=$ (-1)

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Substitution Method

$1)$ $1)$ make one variable the subject
$2)$ $2)$ substitute into second equation
$3)$ $3)$ solve the second equation
$4)$ $4)$ substitute back

First, label the two equations

1

$1$ and

2

$2$ respectively.

$2 m + 5 n$ $2m+5n$	$=$ $=$	$5$ $5$	Equation $1$ $1$
$m - 5 n$ $m-5n$	$=$ $=$	$10$ $10$	Equation $2$ $2$

Next, solve for $m$ $m$ in Equation

2

$2$ .

$m - 5 n$ $m-5n$	$=$ $=$	$10$ $10$
$m - 5 n$ $m-5n$ $+ 5 n$ $+5n$	$=$ $=$	$10$ $10$ $+ 5 n$ $+5n$	Add $5 n$ $5n$ to both sides
$m$ $m$	$=$ $=$	$10 + 5 n$ $10+5n$	Simplify

Substitute $m$ $m$ into Equation

1

$1$ .

$2$ $2$ $m$ $m$ $+ 5 n$ $+5n$	$=$ $=$	$5$ $5$	Equation $1$ $1$
$2$ $2$ $(10 + 5 n)$ $(10+5n)$ $+ 5 n$ $+5n$	$=$ $=$	$5$ $5$	$m = 10 + 5 n$ $m=10+5n$
$20 + 10 n + 5 n$ $20+10n+5n$	$=$ $=$	$5$ $5$	Distribute $2$ $2$ inside the parenthesis
$20 + 15 n$ $20+15n$	$=$ $=$	$5$ $5$	Simplify
$20 + 15 n$ $20+15n$ $- 20$ $-20$	$=$ $=$	$5$ $5$ $- 20$ $-20$	Solve for $n$ $n$
$15 n$ $15n$	$=$ $=$	$- 15$ $-15$
$15 n$ $15n$ $\div 15$ $div15$	$=$ $=$	$- 15$ $-15$ $\div 15$ $div15$	Divide both sides by $15$ $15$
$n$ $n$	$=$ $=$	$- 1$ $-1$

Now, substitute the value of $n$ $n$ into Equation

2

$2$

$m - 5$ $m-5$ $n$ $n$	$=$ $=$	$10$ $10$	Equation $1$ $1$
$m - 5$ $m-5$ $(- 1)$ $(-1)$	$=$ $=$	$10$ $10$	$n = - 1$ $n=-1$
$m + 5$ $m+5$ $- 5$ $-5$	$=$ $=$	$10$ $10$ $- 5$ $-5$	Subtract $5$ $5$ from both sides
$m$ $m$	$=$ $=$	$5$ $5$

m = 5, n = - 1

$m=5, n=-1$

Question 3 of 4

Solve the following systems of equations by substitution.

4 x - 5 y = 14

$4x-5y=14$

2 x + 3 y = - 4

$2x+3y=-4$

$x =$ $x=$ (1)

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

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Substitution Method

$1)$ $1)$ make one variable the subject
$2)$ $2)$ substitute into second equation
$3)$ $3)$ solve the second equation
$4)$ $4)$ substitute back

First, label the two equations

1

$1$ and

2

$2$ respectively.

$4 x - 5 y$ $4x-5y$	$=$ $=$	$14$ $14$	Equation $1$ $1$
$2 x + 3 y$ $2x+3y$	$=$ $=$	$- 4$ $-4$	Equation $2$ $2$

Next, solve for $x$ $x$ in Equation

2

$2$ .

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

$x$ $x$	$=$ $=$	$\frac{- 4 - 3 y}{2}$ $(-4-3y)/2$	Simplify

Substitute $x$ $x$ into Equation

1

$1$ .

$4$ $4$ $x$ $x$ $- 5 y$ $-5y$	$=$ $=$	$14$ $14$	Equation $1$ $1$

$4$ $4$ $(\frac{- 4 - 3 y}{2})$ $((-4-3y)/2)$ $- 5 y$ $-5y$	$=$ $=$	$14$ $14$	$x = \frac{- 4 - 3 y}{2}$ $x=(-4-3y)/2$

$\left(4\left(\frac{-4-3y}{2}\right)\color{#CC0000}{\times2}\right)-(5y\color{#CC0000}{\times2})$	$=$ $=$	$14$ $14$ $\times 2$ $times2$	Multiply all values by $2$ $2$

$4 (- 4 - 3 y) - 10 y$ $4(-4-3y)-10y$	$=$ $=$	$28$ $28$
$- 16 - 12 y - 10 y$ $-16-12y-10y$	$=$ $=$	$28$ $28$	Distribute $4$ $4$ inside the parenthesis
$- 16 - 22 y$ $-16-22y$	$=$ $=$	$28$ $28$	Simplify
$- 16 - 22 y$ $-16-22y$ $+ 16$ $+16$	$=$ $=$	$28$ $28$ $+ 16$ $+16$	Solve for $y$ $y$
$- 22 y$ $-22y$	$=$ $=$	$44$ $44$
$- 22 y$ $-22y$ $\div - 22$ $div-22$	$=$ $=$	$44$ $44$ $\div - 22$ $div-22$	Divide both sides by $- 22$ $-22$
$y$ $y$	$=$ $=$	$- 2$ $-2$

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

2

$2$

$2 x + 3$ $2x+3$ $y$ $y$	$=$ $=$	$- 4$ $-4$	Equation $2$ $2$
$2 x + 3$ $2x+3$ $(- 2)$ $(-2)$	$=$ $=$	$- 4$ $-4$	$y = - 2$ $y=-2$
$2 x - 6$ $2x-6$ $+ 6$ $+6$	$=$ $=$	$- 4$ $-4$ $+ 6$ $+6$	Add $6$ $6$ to both sides
$2 x$ $2x$ $\div 2$ $div2$	$=$ $=$	$2$ $2$ $\div 2$ $div2$	Divide both sides by $2$ $2$
$x$ $x$	$=$ $=$	$1$ $1$

x = 1, y = - 2

$x=1, y=-2$

Question 4 of 4

Solve the following systems of equations by substitution.

7 x - 2 y = 6

$7x-2y=6$

3 x + 4 y = 22

$3x+4y=22$

$x =$ $x=$ (2)

$y =$ $y=$ (4)

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Substitution Method

$1)$ $1)$ make one variable the subject
$2)$ $2)$ substitute into second equation
$3)$ $3)$ solve the second equation
$4)$ $4)$ substitute back

First, label the two equations

1

$1$ and

2

$2$ respectively.

$7 x - 2 y$ $7x-2y$	$=$ $=$	$6$ $6$	Equation $1$ $1$
$3 x + 4 y$ $3x+4y$	$=$ $=$	$22$ $22$	Equation $2$ $2$

Next, solve for $y$ $y$ in Equation

1

$1$ .

$7 x - 2 y$ $7x-2y$	$=$ $=$	$6$ $6$
$7 x - 2 y$ $7x-2y$ $+ 2 y$ $+2y$	$=$ $=$	$6$ $6$ $+ 2 y$ $+2y$	Add $2 y$ $2y$ to both sides
$7 x$ $7x$ $- 6$ $-6$	$=$ $=$	$6 + 2 y$ $6+2y$ $- 6$ $-6$	Subtract $2 y$ $2y$ from both sides
$(7 x - 6)$ $(7x-6)$ $\div 2$ $div2$	$=$ $=$	$2 y$ $2y$ $\div 2$ $div2$	Divide both sides by $2$ $2$

$\frac{7 x - 6}{2}$ $(7x-6)/2$	$=$ $=$	$y$ $y$	Simplify

$y$ $y$	$=$ $=$	$\frac{7 x - 6}{2}$ $(7x-6)/2$	Simplify

Substitute $y$ $y$ into Equation

2

$2$ .

$3 x + 4$ $3x+4$ $y$ $y$	$=$ $=$	$22$ $22$	Equation $2$ $2$

$3 x + 4$ $3x+4$ $(\frac{7 x - 6}{2})$ $((7x-6)/2)$	$=$ $=$	$22$ $22$	$y = \frac{7 x - 6}{2}$ $y=(7x-6)/2$

$(3x\color{#CC0000}{\times2})+\left(4\left(\frac{7x-6}{2}\right)\color{#CC0000}{\times2}\right)$	$=$ $=$	$22$ $22$ $\times 2$ $times2$	Multiply all values by $2$ $2$

$6 x + 4 (7 x - 6)$ $6x+4(7x-6)$	$=$ $=$	$44$ $44$
$6 x + 28 x - 24$ $6x+28x-24$	$=$ $=$	$44$ $44$	Distribute $4$ $4$ inside the parenthesis
$34 x - 24$ $34x-24$	$=$ $=$	$44$ $44$	Simplify
$34 x - 24$ $34x-24$ $+ 24$ $+24$	$=$ $=$	$44$ $44$ $+ 24$ $+24$	Solve for $x$ $x$
$34 x$ $34x$	$=$ $=$	$68$ $68$
$34 x$ $34x$ $\div 34$ $div34$	$=$ $=$	$68$ $68$ $\div 34$ $div34$	Divide both sides by $34$ $34$
$x$ $x$	$=$ $=$	$2$ $2$

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

1

$1$

$7$ $7$ $x$ $x$ $- 2 y$ $-2y$	$=$ $=$	$6$ $6$	Equation $1$ $1$
$7$ $7$ $(2)$ $(2)$ $- 2 y$ $-2y$	$=$ $=$	$6$ $6$	$x = 2$ $x=2$
$14 - 2 y$ $14-2y$ $- 14$ $-14$	$=$ $=$	$6$ $6$ $- 14$ $-14$	Subtract $14$ $14$ from both sides
$- 2 y$ $-2y$ $\div - 2$ $div-2$	$=$ $=$	$- 8$ $-8$ $\div - 2$ $div-2$	Divide both sides by $- 2$ $-2$
$y$ $y$	$=$ $=$	$4$ $4$

x = 2, y = 4

$x=2, y=4$

Substitution Method 4

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Substitution Method

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