Substitution Method 3

Question 1 of 4

Solve the following simultaneous equations by substitution.

8 a + 5 b = 6

$8a+5b=6$

4 a + 2 b = 4

$4a+2b=4$

$a =$ $a=$ (2)

$b =$ $b=$ (-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.

$8 a + 5 b$ $8a+5b$	$=$ $=$	$6$ $6$	Equation $1$ $1$
$4 a + 2 b$ $4a+2b$	$=$ $=$	$4$ $4$	Equation $2$ $2$

Next, solve for $b$ $b$ in Equation

2

$2$ .

$4 a + 2 b$ $4a+2b$	$=$ $=$	$4$ $4$
$(4 a$ $(4a$ $\div 2$ $div2$ $) + (2 b$ $)+(2b$ $\div 2$ $div2$ $)$ $)$	$=$ $=$	$4$ $4$ $\div 2$ $div2$	Divide the values of both sides by $2$ $2$
$2 a + b$ $2a+b$ $- 2 a$ $-2a$	$=$ $=$	$2$ $2$ $- 2 a$ $-2a$	Subtract $2 a$ $2a$ from both sides
$b$ $b$	$=$ $=$	$2 - 2 a$ $2-2a$	Simplify

Substitute $b$ $b$ into Equation

1

$1$ .

$8 a + 5$ $8a+5$ $b$ $b$	$=$ $=$	$6$ $6$	Equation $1$ $1$
$8 a + 5$ $8a+5$ $(2 - 2 a)$ $(2-2a)$	$=$ $=$	$6$ $6$	$b = 2 - 2 a$ $b=2-2a$
$8 a + 10 - 10 a$ $8a+10-10a$	$=$ $=$	$6$ $6$	Distribute $5$ $5$ inside the parenthesis
$10 - 2 a$ $10-2a$	$=$ $=$	$6$ $6$	Simplify
$10 - 2 a$ $10-2a$ $- 10$ $-10$	$=$ $=$	$6$ $6$ $- 10$ $-10$	Solve for $a$ $a$
$- 2 a$ $-2a$	$=$ $=$	$- 4$ $-4$
$- 2 a$ $-2a$ $\div (- 2)$ $div(-2)$	$=$ $=$	$- 4$ $-4$ $\div (- 2)$ $div(-2)$	Divide both sides by $- 2$ $-2$
$a$ $a$	$=$ $=$	$2$ $2$

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

2

$2$

$4$ $4$ $a$ $a$ $+ 2 b$ $+2b$	$=$ $=$	$4$ $4$	Equation $1$ $1$
$4$ $4$ $(2)$ $(2)$ $+ 2 b$ $+2b$	$=$ $=$	$4$ $4$	$a = 2$ $a=2$
$8 + 2 b$ $8+2b$ $- 8$ $-8$	$=$ $=$	$4$ $4$ $- 8$ $-8$	Subtract $8$ $8$ from both sides
$2 b$ $2b$ $\div 2$ $div2$	$=$ $=$	$- 4$ $-4$ $\div 2$ $div2$	Divide both sides by $2$ $2$
$b$ $b$	$=$ $=$	$- 2$ $-2$

a = 2, b = - 2

$a=2, b=-2$

Question 2 of 4

Solve the following simultaneous equations by substitution.

2 x + 5 y = 17

$2x+5y=17$

3 x + 3 y = 12

$3x+3y=12$

$x =$ $x=$ (1)

$y =$ $y=$ (3)

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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.

$2 x + 5 y$ $2x+5y$	$=$ $=$	$17$ $17$	Equation $1$ $1$
$3 x + 3 y$ $3x+3y$	$=$ $=$	$12$ $12$	Equation $2$ $2$

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

2

$2$ .

$3 x + 3 y$ $3x+3y$	$=$ $=$	$12$ $12$
$(3 x$ $(3x$ $\div 3$ $div3$ $) + (3 y$ $)+(3y$ $\div 3$ $div3$ $)$ $)$	$=$ $=$	$12$ $12$ $\div 3$ $div3$	Divide the values of both sides by $3$ $3$
$x + y$ $x+y$ $- y$ $-y$	$=$ $=$	$4$ $4$ $- y$ $-y$	Subtract $y$ $y$ from both sides
$x$ $x$	$=$ $=$	$4 - y$ $4-y$	Simplify

Substitute $x$ $x$ into Equation

1

$1$ .

$2$ $2$ $x$ $x$ $+ 5 y$ $+5y$	$=$ $=$	$17$ $17$	Equation $1$ $1$
$2$ $2$ $(4 - y)$ $(4-y)$ $+ 5 y$ $+5y$	$=$ $=$	$17$ $17$	$x = 4 - y$ $x=4-y$
$8 - 2 y + 5 y$ $8-2y+5y$	$=$ $=$	$17$ $17$	Distribute $2$ $2$ inside the parenthesis
$8 + 3 y$ $8+3y$	$=$ $=$	$17$ $17$	Simplify
$8 + 3 y$ $8+3y$ $- 8$ $-8$	$=$ $=$	$17$ $17$ $- 8$ $-8$	Solve for $y$ $y$
$3 y$ $3y$	$=$ $=$	$9$ $9$
$3 y$ $3y$ $\div 3$ $div3$	$=$ $=$	$9$ $9$ $\div 3$ $div3$	Divide both sides by $3$ $3$
$y$ $y$	$=$ $=$	$3$ $3$

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

2

$2$

$3 x + 3$ $3x+3$ $y$ $y$	$=$ $=$	$12$ $12$	Equation $2$ $2$
$3 x + 3$ $3x+3$ $(3)$ $(3)$	$=$ $=$	$12$ $12$	$y = 3$ $y=3$
$3 x + 9$ $3x+9$ $- 9$ $-9$	$=$ $=$	$12$ $12$ $- 9$ $-9$	Subtract $9$ $9$ to both sides
$3 x$ $3x$ $\div 3$ $div3$	$=$ $=$	$3$ $3$ $\div 3$ $div3$	Divide both sides by $3$ $3$
$x$ $x$	$=$ $=$	$1$ $1$

x = 1, y = 3

$x=1, y=3$

Question 3 of 4

Solve the following simultaneous equations by substitution.

3 x + 7 y = 4

$3x+7y=4$

2 x + 3 y = 1

$2x+3y=1$

$x =$ $x=$ (-1)

$y =$ $y=$ (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.

$3 x + 7 y$ $3x+7y$	$=$ $=$	$4$ $4$	Equation $1$ $1$
$2 x + 3 y$ $2x+3y$	$=$ $=$	$1$ $1$	Equation $2$ $2$

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

2

$2$ .

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

$x$ $x$	$=$ $=$	$\frac{1 - 3 y}{2}$ $(1-3y)/2$	Simplify

Substitute $x$ $x$ into Equation

1

$1$ .

$3$ $3$ $x$ $x$ $+ 7 y$ $+7y$	$=$ $=$	$4$ $4$	Equation $1$ $1$

$3$ $3$ $(\frac{1 - 3 y}{2})$ $((1-3y)/2)$ $+ 7 y$ $+7y$	$=$ $=$	$4$ $4$	$x = \frac{1 - 3 y}{2}$ $x=(1-3y)/2$

$\left(3\left(\frac{1-3y}{2}\right)\color{#CC0000}{\times2}\right)+(7y\color{#CC0000}{\times2})$	$=$ $=$	$4$ $4$ $\times 2$ $times2$	Multiply all values by $2$ $2$

$3 (1 - 3 y) + 14 y$ $3(1-3y)+14y$	$=$ $=$	$8$ $8$
$3 - 9 y + 14 y$ $3-9y+14y$	$=$ $=$	$8$ $8$	Distribute $3$ $3$ inside the parenthesis
$3 + 5 y$ $3+5y$	$=$ $=$	$8$ $8$	Simplify
$3 + 5 y$ $3+5y$ $- 3$ $-3$	$=$ $=$	$8$ $8$ $- 3$ $-3$	Solve for $y$ $y$
$5 y$ $5y$	$=$ $=$	$5$ $5$
$5 y$ $5y$ $\div 5$ $div5$	$=$ $=$	$5$ $5$ $\div 5$ $div5$	Divide both sides by $5$ $5$
$y$ $y$	$=$ $=$	$1$ $1$

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

2

$2$

$2 x + 3$ $2x+3$ $y$ $y$	$=$ $=$	$1$ $1$	Equation $2$ $2$
$2 x + 3$ $2x+3$ $(1)$ $(1)$	$=$ $=$	$1$ $1$	$y = 1$ $y=1$
$2 x + 3$ $2x+3$ $- 3$ $-3$	$=$ $=$	$1$ $1$ $- 3$ $-3$	Subtract $3$ $3$ from 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 = 1

$x=-1, y=1$

Question 4 of 4

Solve the following simultaneous equations by substitution.

5 a + 2 b = 4

$5a+2b=4$

2 a - 3 b = 13

$2a-3b=13$

$a =$ $a=$ (2)

$b =$ $b=$ (-3)

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Mute

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.

$5 a + 2 b$ $5a+2b$	$=$ $=$	$4$ $4$	Equation $1$ $1$
$2 a - 3 b$ $2a-3b$	$=$ $=$	$13$ $13$	Equation $2$ $2$

Next, solve for $a$ $a$ in Equation

2

$2$ .

$2 a - 3 b$ $2a-3b$	$=$ $=$	$13$ $13$
$2 a - 3 b$ $2a-3b$ $+ 3 b$ $+3b$	$=$ $=$	$13$ $13$ $+ 3 b$ $+3b$	Add $3 b$ $3b$ to both sides
$2 a$ $2a$ $\div 2$ $div2$	$=$ $=$	$(13 + 3 b)$ $(13+3b)$ $\div 2$ $div2$	Divide both sides by $2$ $2$

$a$ $a$	$=$ $=$	$\frac{13 + 3 b}{2}$ $(13+3b)/2$	Simplify

Substitute $a$ $a$ into Equation

1

$1$ .

$5$ $5$ $a$ $a$ $+ 2 b$ $+2b$	$=$ $=$	$4$ $4$	Equation $1$ $1$

$5$ $5$ $(\frac{13 + 3 b}{2})$ $((13+3b)/2)$ $+ 2 b$ $+2b$	$=$ $=$	$4$ $4$	$a = \frac{13 + 3 b}{2}$ $a=(13+3b)/2$

$\left(5\left(\frac{13+3b}{2}\right)\color{#CC0000}{\times2}\right)+(2b\color{#CC0000}{\times2})$	$=$ $=$	$4$ $4$ $\times 2$ $times2$	Multiply all values by $2$ $2$

$5 (13 + 3 b) + 4 b$ $5(13+3b)+4b$	$=$ $=$	$8$ $8$
$65 + 15 b + 4 b$ $65+15b+4b$	$=$ $=$	$8$ $8$	Distribute $5$ $5$ inside the parenthesis
$65 + 19 b$ $65+19b$	$=$ $=$	$8$ $8$	Simplify
$65 + 19 b$ $65+19b$ $- 65$ $-65$	$=$ $=$	$8$ $8$ $- 65$ $-65$	Solve for $b$ $b$
$19 b$ $19b$	$=$ $=$	$- 57$ $-57$
$19 b$ $19b$ $\div 19$ $div19$	$=$ $=$	$- 57$ $-57$ $\div 19$ $div19$	Divide both sides by $19$ $19$
$b$ $b$	$=$ $=$	$- 3$ $-3$

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

2

$2$

$2 a - 3$ $2a-3$ $b$ $b$	$=$ $=$	$13$ $13$	Equation $2$ $2$
$2 a - 3$ $2a-3$ $(- 3)$ $(-3)$	$=$ $=$	$13$ $13$	$b = - 3$ $b=-3$
$2 a + 9$ $2a+9$ $- 9$ $-9$	$=$ $=$	$13$ $13$ $- 9$ $-9$	Subtract $9$ $9$ from both sides
$2 a$ $2a$ $\div 2$ $div2$	$=$ $=$	$4$ $4$ $\div 2$ $div2$	Divide both sides by $2$ $2$
$a$ $a$	$=$ $=$	$2$ $2$

a = 2, b = - 3

$a=2, b=-3$

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Substitution Method

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Substitution Method

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Substitution Method

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Substitution Method

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