Equation Problems (Area)

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Question 1 of 5

1. Question
Find $x$ $x$ if the area of the rectangle below is $165$ $165$ cm².
- $x =$ $x=$ (7)
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Inverse Operations

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

Distributive Property

$a$ $a$ $(b + c) =$ $(b+c)=$ $a$ $a$ $b +$ $b+$ $a$ $a$ $c$ $c$

Area of a Rectangle

$A =$ $A=$ $L$ $L$ $\times$ $times$ $W$ $W$

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

$L = (x + 8)$ $L=(x+8)$ cm

$W = 11$ $W=11$ cm

$A = 165$ $A=165$ cm²

$A$ $A$ $=$ $=$ $L$ $L$ $\times$ $times$ $W$ $W$

$165$ $165$ $=$ $=$ $(x + 8)$ $(x+8)$ $\times$ $times$ $11$ $11$ Substitute the values

$165$ $165$ $=$ $=$ $11 (x + 8)$ $11(x+8)$

$11 (x + 8)$ $11(x+8)$ $=$ $=$ $165$ $165$

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

Start by expanding the left side of the equation by using the Distributive Property.

$11$ $11$ $($ $($ $x$ $x$ $+ 8)$ $+8)$ $=$ $=$ $165$ $165$

$11$ $11$ $x$ $x$ $+$ $+$ $11$ $11$ $(8)$ $(8)$ $=$ $=$ $165$ $165$

$11$ $11$ $x$ $x$ $+ 88$ $+88$ $=$ $=$ $165$ $165$

Next, move $88$ $88$ to the other side by subtracting $88$ $88$ from both sides of the equation.

$11$ $11$ $x$ $x$ $+ 88$ $+88$ $=$ $=$ $165$ $165$

$11$ $11$ $x$ $x$ $+ 88$ $+88$ $- 88$ $-88$ $=$ $=$ $165$ $165$ $- 88$ $-88$

$11$ $11$ $x$ $x$ $=$ $=$ $77$ $77$ $88 - 88$ $88-88$ cancels out

Finally, remove $11$ $11$ by dividing both sides of the equation by $11$ $11$ .

$11$ $11$ $x$ $x$ $=$ $=$ $77$ $77$

$11$ $11$ $x$ $x$ $\div 11$ $divide11$ $=$ $=$ $77$ $77$ $\div 11$ $divide11$

$x$ $x$ $=$ $=$ $7$ $7$ $11 \div 11$ $11divide11$ cancels out

$x = 7$ $x=7$
Question 2 of 5

2. Question
Find $y$ $y$ .
- $y =$ $y=$ (4)
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Inverse Operations

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

Distributive Property

$a$ $a$ $(b + c) =$ $(b+c)=$ $a$ $a$ $b +$ $b+$ $a$ $a$ $c$ $c$

Area of a Rectangle

$A =$ $A=$ $L$ $L$ $\times$ $times$ $W$ $W$

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

$L = (6 y - 5)$ $L=(6y-5)$ cm

$W = 10$ $W=10$ cm

$A = 190$ $A=190$ cm²

$A$ $A$ $=$ $=$ $L$ $L$ $\times$ $times$ $W$ $W$

$190$ $190$ $=$ $=$ $(6 y - 5)$ $(6y-5)$ $\times$ $times$ $10$ $10$ Substitute the values

$190$ $190$ $=$ $=$ $10 (6 y - 5)$ $10(6y-5)$

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

Start by expanding the right side of the equation by using the Distributive Property.

$190$ $190$ $=$ $=$ $10$ $10$ $(6$ $(6$ $y$ $y$ $- 5)$ $-5)$

$190$ $190$ $=$ $=$ $10$ $10$ $(6$ $(6$ $y$ $y$ $) +$ $)+$ $10$ $10$ $(- 5)$ $(-5)$

$190$ $190$ $=$ $=$ $60$ $60$ $y$ $y$ $- 50$ $-50$

Next, move $50$ $50$ to the other side by adding $50$ $50$ to both sides of the equation.

$190$ $190$ $=$ $=$ $60$ $60$ $y$ $y$ $- 50$ $-50$

$190$ $190$ $+ 50$ $+50$ $=$ $=$ $60$ $60$ $y$ $y$ $- 50$ $-50$ $+ 50$ $+50$

$240$ $240$ $=$ $=$ $60$ $60$ $y$ $y$ $- 50 + 50$ $-50+50$ cancels out

Finally, remove $60$ $60$ by dividing both sides of the equation by $60$ $60$ .

$240$ $240$ $=$ $=$ $60$ $60$ $y$ $y$

$240$ $240$ $\div 60$ $divide60$ $=$ $=$ $60$ $60$ $y$ $y$ $\div 60$ $divide60$

$4$ $4$ $=$ $=$ $y$ $y$ $60 \div 60$ $60divide60$ cancels out

$y$ $y$ $=$ $=$ $4$ $4$

$y = 4$ $y=4$
Question 3 of 5

3. Question
Find $n$ $n$ if the area of the triangle below is $140$ $140$ cm².
- $n =$ $n=$ (13)
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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

$A = \frac{1}{2}$ $A=1/2$ $b$ $b$ $h$ $h$

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

$b = (2 n + 9)$ $b=(2n+9)$ cm

$h = 8$ $h=8$ cm

$A = 140$ $A=140$ cm²

$A$ $A$ $=$ $=$ $\frac{1}{2}$ $1/2$ $b$ $b$ $h$ $h$

$140$ $140$ $=$ $=$ $\frac{1}{2}$ $1/2$ $(2 n + 9)$ $(2n+9)$ $8$ $8$ Substitute the values

$140$ $140$ $=$ $=$ $4 (2 n + 9)$ $4(2n+9)$ Simplify

$4 (2 n + 9)$ $4(2n+9)$ $=$ $=$ $140$ $140$

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

Start with removing $4$ $4$ by dividing both sides of the equation by $4$ $4$ .

$4 (2$ $4(2$ $n$ $n$ $+ 9)$ $+9)$ $=$ $=$ $140$ $140$

$4 (2$ $4(2$ $n$ $n$ $+ 9)$ $+9)$ $\div 4$ $divide4$ $=$ $=$ $140$ $140$ $\div 4$ $divide4$

$2$ $2$ $n$ $n$ $+ 9$ $+9$ $=$ $=$ $35$ $35$ $4 \div 4$ $4-:4$ cancels out

Next, move $9$ $9$ to the other side by subtracting $9$ $9$ from both sides of the equation.

$2$ $2$ $n$ $n$ $+ 9$ $+9$ $=$ $=$ $35$ $35$

$2$ $2$ $n$ $n$ $+ 9$ $+9$ $- 9$ $-9$ $=$ $=$ $35$ $35$ $- 9$ $-9$

$2$ $2$ $n$ $n$ $=$ $=$ $26$ $26$ $9 - 9$ $9-9$ cancels out

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

$2$ $2$ $n$ $n$ $=$ $=$ $26$ $26$

$2$ $2$ $n$ $n$ $\div 2$ $divide2$ $=$ $=$ $26$ $26$ $\div 2$ $divide2$

$n$ $n$ $=$ $=$ $13$ $13$ $2 \div 2$ $2divide2$ cancels out

$n = 13$ $n=13$
Question 4 of 5

4. Question
Find $h$ $h$ .
- $h =$ $h=$ (20)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

$A = \frac{1}{2}$ $A=1/2$ $b$ $b$ $h$ $h$

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

$A = 170$ $A=170$ cm²

$b = 17$ $b=17$ cm

$h = ?$ $h=?$ cm

$A$ $A$ $=$ $=$ $\frac{1}{2}$ $1/2$ $b$ $h$

$170$ $=$ $1/2$ $(17)$ $h$ Substitute the values

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

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

$170$ $=$ $1/2 (17)$ $h$

$170$ $times2$ $=$ $1/2 (17)$ $h$ $times2$

$340$ $=$ $17$ $h$ $1/2times2$ cancels out

Finally, remove $17$ by dividing both sides of the equation by $17$ .

$340$ $=$ $17$ $h$

$340$ $divide17$ $=$ $17$ $h$ $divide17$

$20$ $=$ $h$ $17divide17$ cancels out

$h$ $=$ $20$ cm

$h=20$ cm
Question 5 of 5

5. Question
Find the dimensions and the area of the rectangle below.
- $L=$ (69)cm
  
  $W=$ (7)cm
  
  $A=$ (483)cm²
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Inverse Operations

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

Distributive Property

$a$ $(b+c)=$ $a$ $b+$ $a$ $c$

Area of a Rectangle

$A=$ $L$ $times$ $W$

First, find the value of $x$ .

Do this by forming an equation knowing that opposite sides of a rectangle are equal.

$3x+6$ $=$ $4x-15$

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

Start by moving $3x$ to the other side by subtracting $3x$ from both sides of the equation.

$3$ $x$ $+6$ $=$ $4$ $x$ $-15$

$3$ $x$ $+6$ $-3x$ $=$ $4$ $x$ $-15$ $-3x$

$6$ $=$ $x$ $-15$ $3x-3x$ cancels out

Next, move $15$ to the other side by adding $15$ to both sides of the equation.

$6$ $=$ $x$ $-15$

$6$ $+15$ $=$ $x$ $-15$ $+15$

$21$ $=$ $x$ $-15+15$ cancels out

$x$ $=$ $21$

Substitute $x=21$ to find the dimensions of the rectangle.

Width $=$ $1/3$ $x$

$=$ $1/3$ $(21)$ cm

$=$ $7$ cm

Length $=$ $3$ $x$ $+6$

$=$ $3$ $(21)$ $+6$

$=$ $63+6$

$=$ $69$ cm

Finally, use the dimensions of the rectangle to find its Area.

$L=69$ cm

$W=7$ cm

$A$ $=$ $L$ $times$ $W$ Area Formula

$=$ $69$ $times$ $7$ Substitute the values

$=$ $483$ cm² Simplify

$L=69$ cm

$W=7$ cm

$A=483$ cm²

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Quiz summary

Information

1. Question

Hint

Well Done!

Inverse Operations

Distributive Property

Area of a Rectangle

2. Question

Hint

Fantastic!

Inverse Operations

Distributive Property

Area of a Rectangle

3. Question

Hint

Good Job!

Inverse Operations

Area of a Triangle

4. Question

Hint

Excellent!

Inverse Operations

Area of a Triangle

5. Question

Hint

Exceptional!

Inverse Operations

Distributive Property

Area of a Rectangle

Quizzes

$A$ $A$	$=$ $=$	$L$ $L$ $\times$ $times$ $W$ $W$
$165$ $165$	$=$ $=$	$(x + 8)$ $(x+8)$ $\times$ $times$ $11$ $11$	Substitute the values
$165$ $165$	$=$ $=$	$11 (x + 8)$ $11(x+8)$
$11 (x + 8)$ $11(x+8)$	$=$ $=$	$165$ $165$

$11$ $11$ $($ $($ $x$ $x$ $+ 8)$ $+8)$	$=$ $=$	$165$ $165$
$11$ $11$ $x$ $x$ $+$ $+$ $11$ $11$ $(8)$ $(8)$	$=$ $=$	$165$ $165$
$11$ $11$ $x$ $x$ $+ 88$ $+88$	$=$ $=$	$165$ $165$

$11$ $11$ $x$ $x$	$=$ $=$	$77$ $77$
$11$ $11$ $x$ $x$ $\div 11$ $divide11$	$=$ $=$	$77$ $77$ $\div 11$ $divide11$
$x$ $x$	$=$ $=$	$7$ $7$	$11 \div 11$ $11divide11$ cancels out

$190$ $190$	$=$ $=$	$10$ $10$ $(6$ $(6$ $y$ $y$ $- 5)$ $-5)$
$190$ $190$	$=$ $=$	$10$ $10$ $(6$ $(6$ $y$ $y$ $) +$ $)+$ $10$ $10$ $(- 5)$ $(-5)$
$190$ $190$	$=$ $=$	$60$ $60$ $y$ $y$ $- 50$ $-50$

$240$ $240$	$=$ $=$	$60$ $60$ $y$ $y$
$240$ $240$ $\div 60$ $divide60$	$=$ $=$	$60$ $60$ $y$ $y$ $\div 60$ $divide60$
$4$ $4$	$=$ $=$	$y$ $y$	$60 \div 60$ $60divide60$ cancels out
$y$ $y$	$=$ $=$	$4$ $4$

$A$ $A$	$=$ $=$	$\frac{1}{2}$ $1/2$ $b$ $b$ $h$ $h$

$140$ $140$	$=$ $=$	$\frac{1}{2}$ $1/2$ $(2 n + 9)$ $(2n+9)$ $8$ $8$	Substitute the values

$140$ $140$	$=$ $=$	$4 (2 n + 9)$ $4(2n+9)$	Simplify
$4 (2 n + 9)$ $4(2n+9)$	$=$ $=$	$140$ $140$

$4 (2$ $4(2$ $n$ $n$ $+ 9)$ $+9)$	$=$ $=$	$140$ $140$
$4 (2$ $4(2$ $n$ $n$ $+ 9)$ $+9)$ $\div 4$ $divide4$	$=$ $=$	$140$ $140$ $\div 4$ $divide4$
$2$ $2$ $n$ $n$ $+ 9$ $+9$	$=$ $=$	$35$ $35$	$4 \div 4$ $4-:4$ cancels out

$2$ $2$ $n$ $n$	$=$ $=$	$26$ $26$
$2$ $2$ $n$ $n$ $\div 2$ $divide2$	$=$ $=$	$26$ $26$ $\div 2$ $divide2$
$n$ $n$	$=$ $=$	$13$ $13$	$2 \div 2$ $2divide2$ cancels out

$A$ $A$	$=$ $=$	$\frac{1}{2}$ $1/2$ $b$ $h$

$170$	$=$	$1/2$ $(17)$ $h$	Substitute the values

$170$	$=$	$1/2 (17)$ $h$

$170$ $times2$	$=$	$1/2 (17)$ $h$ $times2$

$340$	$=$	$17$ $h$	$1/2times2$ cancels out

$3$ $x$ $+6$	$=$	$4$ $x$ $-15$
$3$ $x$ $+6$ $-3x$	$=$	$4$ $x$ $-15$ $-3x$
$6$	$=$	$x$ $-15$	$3x-3x$ cancels out

$6$	$=$	$x$ $-15$
$6$ $+15$	$=$	$x$ $-15$ $+15$
$21$	$=$	$x$	$-15+15$ cancels out
$x$	$=$	$21$

Width	$=$	$1/3$ $x$

	$=$	$1/3$ $(21)$ cm

	$=$	$7$ cm

Length	$=$	$3$ $x$ $+6$
	$=$	$3$ $(21)$ $+6$
	$=$	$63+6$
	$=$	$69$ cm

$340$	$=$	$17$ $h$
$340$ $divide17$	$=$	$17$ $h$ $divide17$
$20$	$=$	$h$	$17divide17$ cancels out
$h$	$=$	$20$ cm

$A$	$=$	$L$ $times$ $W$	Area Formula
	$=$	$69$ $times$ $7$	Substitute the values
	$=$	$483$ cm²	Simplify