Substitution Method 1

Question 1 of 5

Solve the following simultaneous equations by substitution.

x - y = 8

$x-y=8$

2 x + y = 1

$2x+y=1$

$x =$ $x=$ (3)

$y =$ $y=$ (-5)

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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.

$x - y$ $x-y$	$=$ $=$	$8$ $8$	Equation $1$ $1$
$2 x + y$ $2x+y$	$=$ $=$	$1$ $1$	Equation $2$ $2$

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

1

$1$ .

$x - y$ $x-y$	$=$ $=$	$8$ $8$
$x - y$ $x-y$ $+ y$ $+y$	$=$ $=$	$8$ $8$ $+ y$ $+y$	Add $y$ $y$ to both sides
$x$ $x$	$=$ $=$	$8 + y$ $8+y$	Simplify

Substitute $x$ $x$ into Equation

2

$2$ .

$2$ $2$ $x$ $x$ $+ y$ $+y$	$=$ $=$	$1$ $1$	Equation $2$ $2$
$2$ $2$ $(8 + y)$ $(8+y)$ $+ y$ $+y$	$=$ $=$	$1$ $1$	$x = 8 + y$ $x=8+y$
$16 + 2 y + y$ $16+2y+y$	$=$ $=$	$1$ $1$	Distribute $2$ $2$ inside the parenthesis
$16 + 3 y$ $16+3y$	$=$ $=$	$1$ $1$	Simplify
$16 + 3 y$ $16+3y$ $- 16$ $-16$	$=$ $=$	$1$ $1$ $- 16$ $-16$	Solve for $y$ $y$
$3 y$ $3y$	$=$ $=$	$- 15$ $-15$
$y$ $y$	$=$ $=$	$- 5$ $-5$	Divide both sides by $3$ $3$

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

1

$1$

$x -$ $x-$ $y$ $y$	$=$ $=$	$8$ $8$	Equation $1$ $1$
$x -$ $x-$ $(- 5)$ $(-5)$	$=$ $=$	$8$ $8$	$y = - 5$ $y=-5$
$x + 5$ $x+5$	$=$ $=$	$8$ $8$
$x + 5$ $x+5$ $- 5$ $-5$	$=$ $=$	$8$ $8$ $- 5$ $-5$	Simplify
$x$ $x$	$=$ $=$	$3$ $3$

x = 3, y = - 5

$x=3, y =-5$

Question 2 of 5

Solve the following simultaneous equations by substitution.

2 a + b = 5

$2a+b=5$

5 a - 3 b = 7

$5a-3b=7$

$a =$ $a=$ (2)

$b =$ $b=$ (1)

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The substitution method removes one of the variables by replacement in a pair of systems of equations.

First, label the two equations

1

$1$ and

2

$2$ respectively.

$2 a + b$ $2a+b$	$=$ $=$	$5$ $5$	Equation $1$ $1$
$5 a - 3 b$ $5a-3b$	$=$ $=$	$7$ $7$	Equation $2$ $2$

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

1

$1$ .

$2 a + b$ $2a+b$	$=$ $=$	$5$ $5$
$2 a + b$ $2a+b$ $- 2 a$ $-2a$	$=$ $=$	$5$ $5$ $- 2 a$ $-2a$	Subtract $2 a$ $2a$ from both sides
$b$ $b$	$=$ $=$	$5 - 2 a$ $5-2a$	Simplify

Substitute $b$ $b$ into Equation

2

$2$ .

$5 a - 3$ $5a-3$ $b$ $b$	$=$ $=$	$7$ $7$	Equation $2$ $2$
$5 a - 3$ $5a-3$ $(5 - 2 a)$ $(5-2a)$	$=$ $=$	$7$ $7$	$b = 5 - 2 a$ $b=5-2a$
$5 a - 15 + 6 a$ $5a-15+6a$	$=$ $=$	$7$ $7$	Distribute $- 3$ $-3$ inside the parenthesis
$11 a - 15$ $11a-15$	$=$ $=$	$7$ $7$	Simplify
$11 a - 15$ $11a-15$ $+ 15$ $+15$	$=$ $=$	$7$ $7$ $+ 15$ $+15$	Solve for $a$ $a$
$11 a$ $11a$	$=$ $=$	$22$ $22$
$a$ $a$	$=$ $=$	$2$ $2$	Divide both sides by $11$ $11$

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

1

$1$

$2$ $2$ $a$ $a$ $+ b$ $+b$	$=$ $=$	$5$ $5$	Equation $1$ $1$
$2$ $2$ $(2)$ $(2)$ $+ b$ $+b$	$=$ $=$	$5$ $5$	$a = 2$ $a=2$
$4 + b$ $4+b$	$=$ $=$	$5$ $5$
$4 + b$ $4+b$ $- 4$ $-4$	$=$ $=$	$5$ $5$ $- 4$ $-4$	Simplify
$b$ $b$	$=$ $=$	$1$ $1$

a = 2, b = 1

$a=2, b=1$

Question 3 of 5

Solve the following simultaneous equations by substitution.

x - y = 1

$x-y=1$

2 x + y = 11

$2x+y=11$

$x =$ $x=$ (4)

$y =$ $y=$ (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.

$x - y$ $x-y$	$=$ $=$	$1$ $1$	Equation $1$ $1$
$2 x + y$ $2x+y$	$=$ $=$	$11$ $11$	Equation $2$ $2$

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

1

$1$ .

$x - y$ $x-y$	$=$ $=$	$1$ $1$
$x - y$ $x-y$ $+ y$ $+y$	$=$ $=$	$1$ $1$ $+ y$ $+y$	Add $y$ $y$ to both sides
$x$ $x$	$=$ $=$	$1 + y$ $1+y$	Simplify

Substitute $x$ $x$ into Equation

2

$2$ .

$2$ $2$ $x$ $x$ $+ y$ $+y$	$=$ $=$	$11$ $11$	Equation $2$ $2$
$2$ $2$ $(1 + y)$ $(1+y)$ $+ y$ $+y$	$=$ $=$	$11$ $11$	$x = 1 + y$ $x=1+y$
$2 y + 2 + y$ $2y+2+y$	$=$ $=$	$11$ $11$	Distribute $2$ $2$ inside the parenthesis
$3 y + 2$ $3y+2$	$=$ $=$	$11$ $11$	Simplify
$3 y + 2$ $3y+2$ $- 2$ $-2$	$=$ $=$	$11$ $11$ $- 2$ $-2$	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

1

$1$

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

x = 4, y = 3

$x=4, y =3$

Question 4 of 5

Solve the following simultaneous equations by substitution.

a - 3 b = 2

$a-3b=2$

2 a - b = 14

$2a-b=14$

$a =$ $a=$ (8)

$b =$ $b=$ (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.

$a - 3 b$ $a-3b$	$=$ $=$	$2$ $2$	Equation $1$ $1$
$2 a - b$ $2a-b$	$=$ $=$	$14$ $14$	Equation $2$ $2$

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

1

$1$ .

$a - 3 b$ $a-3b$	$=$ $=$	$2$ $2$
$a - 3 b$ $a-3b$ $+ 3 b$ $+3b$	$=$ $=$	$2$ $2$ $+ 3 b$ $+3b$	Add $3 b$ $3b$ to both sides
$a$ $a$	$=$ $=$	$2 + 3 b$ $2+3b$	Simplify

Substitute $a$ $a$ into Equation

2

$2$ .

$2$ $2$ $a$ $a$ $- b$ $-b$	$=$ $=$	$14$ $14$	Equation $2$ $2$
$2$ $2$ $(2 + 3 b)$ $(2+3b)$ $- b$ $-b$	$=$ $=$	$14$ $14$	$a = 2 + 3 b$ $a=2+3b$
$4 + 6 b - b$ $4+6b-b$	$=$ $=$	$14$ $14$	Distribute $2$ $2$ inside the parenthesis
$5 b + 4$ $5b+4$	$=$ $=$	$14$ $14$	Simplify
$5 b + 4$ $5b+4$ $- 4$ $-4$	$=$ $=$	$14$ $14$ $- 4$ $-4$	Solve for $b$ $b$
$5 b$ $5b$	$=$ $=$	$10$ $10$
$5 b$ $5b$ $\div 5$ $div5$	$=$ $=$	$10$ $10$ $\div 5$ $div5$	Divide both sides by $5$ $5$
$b$ $b$	$=$ $=$	$2$ $2$

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

1

$1$

$a - 3$ $a-3$ $b$ $b$	$=$ $=$	$2$ $2$	Equation $1$ $1$
$a - 3$ $a-3$ $(2)$ $(2)$	$=$ $=$	$2$ $2$	$b = 2$ $b=2$
$a - 6$ $a-6$ $+ 6$ $+6$	$=$ $=$	$2$ $2$ $+ 6$ $+6$	Add $6$ $6$ to both sides
$a$ $a$	$=$ $=$	$8$ $8$

a = 8, b = 2

$a=8, b=2$

Question 5 of 5

Solve the following simultaneous equations by substitution.

2 x + 5 y = 7

$2x+5y=7$

x - y = 7

$x-y=7$

$x =$ $x=$ (6)

$y =$ $y=$ (-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

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.

$2 x + 5 y$ $2x+5y$	$=$ $=$	$7$ $7$	Equation $1$ $1$
$x - y$ $x-y$	$=$ $=$	$7$ $7$	Equation $2$ $2$

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

2

$2$ .

$x - y$ $x-y$	$=$ $=$	$7$ $7$
$x - y$ $x-y$ $+ y$ $+y$	$=$ $=$	$7$ $7$ $+ y$ $+y$	Add $y$ $y$ from both sides
$x$ $x$	$=$ $=$	$7 + y$ $7+y$	Simplify

Substitute $x$ $x$ into Equation

1

$1$ .

$2$ $2$ $x$ $x$ $+ 5 y$ $+5y$	$=$ $=$	$7$ $7$	Equation $1$ $1$
$2$ $2$ $(7 + y)$ $(7+y)$ $+ 5 y$ $+5y$	$=$ $=$	$7$ $7$	$x = 7 + y$ $x=7+y$
$14 + 2 y + 5 y$ $14+2y+5y$	$=$ $=$	$7$ $7$	Distribute $2$ $2$ inside the parenthesis
$14 + 7 y$ $14+7y$	$=$ $=$	$7$ $7$	Simplify
$14 + 7 y$ $14+7y$ $- 14$ $-14$	$=$ $=$	$7$ $7$ $- 14$ $-14$	Solve for $y$ $y$
$7 y$ $7y$	$=$ $=$	$- 7$ $-7$
$7 y$ $7y$ $\div 7$ $div7$	$=$ $=$	$- 7$ $-7$ $\div 7$ $div7$	Divide both sides by $7$ $7$
$y$ $y$	$=$ $=$	$- 1$ $-1$

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

2

$2$

$x -$ $x-$ $y$ $y$	$=$ $=$	$7$ $7$	Equation $2$ $2$
$x -$ $x-$ $(- 1)$ $(-1)$	$=$ $=$	$7$ $7$	$y = - 2$ $y=-2$
$x + 1$ $x+1$	$=$ $=$	$7$ $7$	Simplify
$x + 1$ $x+1$ $- 1$ $-1$	$=$ $=$	$7$ $7$ $- 1$ $-1$	Subtract $1$ $1$ from both sides
$x$ $x$	$=$ $=$	$6$ $6$

x = 6, y = - 1

$x=6, y=-1$

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Quiz summary

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1. Question

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Great Work!

Substitution Method

2. Question

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3. Question

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Great Work!

Substitution Method

4. Question

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Keep Going!

Substitution Method

5. Question

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Excellent!

Substitution Method

Substitution Method

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