Equation Problems with Substitution

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Question 1 of 6

1. Question
Find $w$ if the perimeter of the rectangle below is $22$ cm.
- $w=$ (4)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

Perimeter of a Rectangle

$P=2$ $L$ $+2$ $W$

Form an equation using the formula for the Perimeter of a Rectangle.

$P=22$ cm

$L=7$ cm

$w$ cm

$P$ $=$ $2$ $L$ $+2$ $W$

$22$ $=$ $2($ $7$ $)+2$ $w$ Substitute the values

$22$ $=$ $14+2w$ Simplify

To solve for $w$ , it needs to be alone on one side.

Start by moving $14$ to the other side by subtracting $14$ from both sides of the equation.

$22$ $=$ $14+2w$

$22$ $-14$ $=$ $14+2w$ $-14$

$8$ $=$ $2$ $w$ $14-14$ cancels out

Finally, remove $2$ by dividing both sides of the equation by $2$ .

$8$ $=$ $2$ $w$

$8$ $divide2$ $=$ $2$ $w$ $divide2$

$4$ $=$ $w$

$w$ $=$ $4$

Check our work

To confirm our answer, substitute $w=4$ to the formed equation.

$22$ $=$ $14+2w$

$22$ $=$ $14+2(4)$ Substitute $w=4$

$22$ $=$ $14+8$

$22$ $=$ $22$

Since the equation is true, the answer is correct.

$w=4$
Question 2 of 6

2. Question
Find the height $(h)$ of the triangle below if its area is $45$ cm².
- (9) cm
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

Area of a Triangle

$A=1/2$ $b$ $h$

First, label the values and form an equation using the Area of a Triangle formula.

$b=10$ cm

$h=?$ cm

$A=45$ cm²

$A$ $=$ $1/2$ $b$ $h$

$45$ $=$ $1/2$ $10$ $h$ Substitute the values

$45$ $=$ $5h$ Simplify

To solve for $h$ , it needs to be alone on one side.

Remove $5$ by dividing both sides of the equation by $5$ .

$45$ $=$ $5$ $h$

$45$ $divide5$ $=$ $5$ $h$ $divide5$

$9$ $=$ $h$ $5-:5$ cancels out

$h$ $=$ $9$ cm

$9$ cm
Question 3 of 6

3. Question
Find the height $(h)$ of the trapezium below if $a=7$ , $b=9$ and its area $(A)$ is $72$ .
- $h=$ (9)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

Area of a Trapezium

$A=1/2$ $h$ $($ $a$ $+$ $b$ $)$

First, label the values and form an equation using the Area of a Trapezium formula.

$A=72$

$a=7$

$b=9$

$h=?$

$A$ $=$ $1/2$ $h$ $($ $a$ $+$ $b$ $)$

$72$ $=$ $1/2$ $h$ $($ $7$ $+$ $9$ $)$ Substitute the values

$72$ $=$ $1/2 h(16)$ Simplify

$72$ $=$ $8h$

To solve for $h$ , it needs to be alone on one side.

Remove $8$ by dividing both sides of the equation by $8$ .

$72$ $=$ $8$ $h$

$72$ $divide8$ $=$ $8$ $h$ $divide8$

$9$ $=$ $h$ $8-:8$ cancels out

$h$ $=$ $9$

$h=9$
Question 4 of 6

4. Question
Given that $V=u+at$ , find $a$ using the following values:

$V=178$

$u=28$

$t=10$
- $a=$ (15)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

First, list the values and form an equation using the given formula for $V$ .

$V=178$

$u=28$

$t=10$

$V$ $=$ $u$ $+a$ $t$

$178$ $=$ $28$ $+a$ $(10)$ Substitute the values

$178$ $=$ $28+10a$ Simplify

To solve for $a$ , it needs to be alone on one side.

Start by moving $28$ to the other side by subtracting $28$ from both sides of the equation.

$178$ $=$ $28+10$ $a$

$178$ $-28$ $=$ $28+10$ $a$ $-28$

$150$ $=$ $10$ $a$ $28-28$ cancels out

Finally, remove $10$ by dividing both sides of the equation by $10$ .

$150$ $=$ $10$ $a$

$150$ $divide10$ $=$ $10$ $a$ $divide10$

$15$ $=$ $a$ $10divide10$ cancels out

$a$ $=$ $15$

$a=15$
Question 5 of 6

5. Question
Given that $A=(x+y)/2$ , find $y$ using the following values:

$A=51$

$x=67$
- $y=$ (35)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

First, list the values and form an equation using the given formula for $A$ .

$A=51$

$x=67$

$A$ $=$ $\frac{\color{#007DDC}{x}+y}{2}$

$51$ $=$ $\frac{\color{#007DDC}{67}+y}{2}$ Substitute the values

To solve for $y$ , it needs to be alone on one side.

Start by removing $1/2$ by multiplying both sides of the equation by $2$ .

$51$ $=$ $\frac{67+\color{#00880A}{y}}{2}$

$51$ $times2$ $=$ $\left(\frac{67+\color{#00880A}{y}}{2}\right)\color{#CC0000}{\times2}$

$102$ $=$ $67+$ $y$ $1/2times2$ cancels out

Finally, move $67$ to the other side by subtracting $67$ from both sides of the equation.

$102$ $=$ $67+$ $y$

$102$ $-67$ $=$ $67+$ $y$ $-67$

$35$ $=$ $y$ $67-67$ cancels out

$y$ $=$ $35$

Check our work

To confirm our answer, substitute $y=35$ to the original equation.

$51$ $=$ $(67+y)/2$

$51$ $=$ $(67+35)/2$ Substitute $y=35$

$51$ $=$ $102/2$

$51$ $=$ $51$

Since the equation is true, the answer is correct.

$y=35$
Question 6 of 6

6. Question
Convert $30°C$ to $F$ using the formula below:

$C=5/9 (F-32)$
- (86) $°F$
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

First, use the formula and the given value to form an equation.

$C=30°$

$C$ $=$ $5/9 (F-32)$

$30$ $=$ $5/9 (F-32)$ Substitute $C$

To solve for $F$ , it needs to be alone on one side.

Start by removing $1/9$ by multiplying both sides of the equation by $9$ .

$30$ $=$ $5/9 ($ $F$ $-32)$

$30$ $times9$ $=$ $5/9 ($ $F$ $-32)$ $times9$

$270$ $=$ $5($ $F$ $-32)$ $1/9times9$ cancels out

Next, remove $5$ by dividing both sides of the equation by $5$ .

$270$ $=$ $5($ $F$ $-32)$

$270$ $divide5$ $=$ $5($ $F$ $-32)$ $divide5$

$54$ $=$ $F$ $-32$ $5divide5$ cancels out

Finally, move $32$ to the other side by adding $32$ to both sides of the equation.

$54$ $=$ $F$ $-32$

$54$ $+32$ $=$ $F$ $-32$ $+32$

$86$ $=$ $F$ $-32+32$ cancels out

$F$ $=$ $86°$

Therefore, $30°C$ is equal to $86°F$

$86°F$

Equation Problems with Substitution

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Quiz summary

Information

1. Question

Hint

Good Job!

Inverse Operations

Perimeter of a Rectangle

2. Question

Hint

Excellent!

Inverse Operations

Area of a Triangle

3. Question

Hint

Great Work!

Inverse Operations

Area of a Trapezium

4. Question

Hint

Fantastic!

Inverse Operations

5. Question

Hint

Well Done!

Inverse Operations

6. Question

Hint

Excellent!

Inverse Operations

Quizzes

$P$	$=$	$2$ $L$ $+2$ $W$
$22$	$=$	$2($ $7$ $)+2$ $w$	Substitute the values
$22$	$=$	$14+2w$	Simplify

$22$	$=$	$14+2w$
$22$ $-14$	$=$	$14+2w$ $-14$
$8$	$=$	$2$ $w$	$14-14$ cancels out

$8$	$=$	$2$ $w$
$8$ $divide2$	$=$	$2$ $w$ $divide2$
$4$	$=$	$w$
$w$	$=$	$4$

$A$	$=$	$1/2$ $b$ $h$

$45$	$=$	$1/2$ $10$ $h$	Substitute the values

$45$	$=$	$5h$	Simplify

$45$	$=$	$5$ $h$
$45$ $divide5$	$=$	$5$ $h$ $divide5$
$9$	$=$	$h$	$5-:5$ cancels out
$h$	$=$	$9$ cm

$A$	$=$	$1/2$ $h$ $($ $a$ $+$ $b$ $)$

$72$	$=$	$1/2$ $h$ $($ $7$ $+$ $9$ $)$	Substitute the values

$72$	$=$	$1/2 h(16)$	Simplify

$72$	$=$	$8h$

$72$	$=$	$8$ $h$
$72$ $divide8$	$=$	$8$ $h$ $divide8$
$9$	$=$	$h$	$8-:8$ cancels out
$h$	$=$	$9$

$V$	$=$	$u$ $+a$ $t$
$178$	$=$	$28$ $+a$ $(10)$	Substitute the values
$178$	$=$	$28+10a$	Simplify

$178$	$=$	$28+10$ $a$
$178$ $-28$	$=$	$28+10$ $a$ $-28$
$150$	$=$	$10$ $a$	$28-28$ cancels out

$150$	$=$	$10$ $a$
$150$ $divide10$	$=$	$10$ $a$ $divide10$
$15$	$=$	$a$	$10divide10$ cancels out
$a$	$=$	$15$

$51$	$=$	$\frac{67+\color{#00880A}{y}}{2}$

$51$ $times2$	$=$	$\left(\frac{67+\color{#00880A}{y}}{2}\right)\color{#CC0000}{\times2}$

$102$	$=$	$67+$ $y$	$1/2times2$ cancels out

$102$	$=$	$67+$ $y$
$102$ $-67$	$=$	$67+$ $y$ $-67$
$35$	$=$	$y$	$67-67$ cancels out
$y$	$=$	$35$

$51$	$=$	$(67+y)/2$

$51$	$=$	$(67+35)/2$	Substitute $y=35$

$51$	$=$	$102/2$

$51$	$=$	$51$

$C$	$=$	$5/9 (F-32)$

$30$	$=$	$5/9 (F-32)$	Substitute $C$

$A$	$=$	$\frac{\color{#007DDC}{x}+y}{2}$

$51$	$=$	$\frac{\color{#007DDC}{67}+y}{2}$	Substitute the values

$30$	$=$	$5/9 ($ $F$ $-32)$

$30$ $times9$	$=$	$5/9 ($ $F$ $-32)$ $times9$

$270$	$=$	$5($ $F$ $-32)$	$1/9times9$ cancels out

$270$	$=$	$5($ $F$ $-32)$
$270$ $divide5$	$=$	$5($ $F$ $-32)$ $divide5$
$54$	$=$	$F$ $-32$	$5divide5$ cancels out

$54$	$=$	$F$ $-32$
$54$ $+32$	$=$	$F$ $-32$ $+32$
$86$	$=$	$F$	$-32+32$ cancels out
$F$	$=$	$86°$