Equation Problems (Geometry) 2

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Year 8

Equations

Equation Problems (Geometry)

Equation Problems (Geometry) 2

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Question 1 of 4

1. Question
Find $y$
- $y=$ (25)
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Inverse Operations

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

Form an equation knowing that alternate angles are equal.

$3y+15$ $=$ $2y+40$

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

Start by moving $2y$ to the other side by subtracting $2y$ from both sides of the equation.

$3$ $y$ $+15$ $=$ $2$ $y$ $+40$

$3$ $y$ $+15$ $-2y$ $=$ $2$ $y$ $+40$ $-2y$

$y$ $+15$ $=$ $40$ $2y-2y$ cancels out

Finally, move $15$ to the other side by subtracting $15$ from both sides of the equation.

$y$ $+15$ $=$ $40$

$y$ $+15$ $-15$ $=$ $40$ $-15$

$y$ $=$ $25$ $15-15$ cancels out

Check our work

To confirm our answer, substitute $y=25$ to the formed equation.

$3y+15$ $=$ $2y+40$

$3(25)+15$ $=$ $2(25)+40$ Substitute $y=25$

$75+15$ $=$ $50+40$

$90$ $=$ $90$

Since the equation is true, the answer is correct.

$y=25$
Question 2 of 4

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

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

Form an equation knowing that corresponding angles are equal.

$7y-27$ $=$ $5y-19$

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

Start by moving $5y$ to the other side by subtracting $5y$ from both sides of the equation.

$7$ $y$ $-27$ $=$ $5$ $y$ $-19$

$7$ $y$ $-27$ $-5y$ $=$ $5$ $y$ $-19$ $-5y$

$2$ $y$ $-27$ $=$ $-19$ $5y-5y$ cancels out

Next, move $27$ to the other side by adding $27$ to both sides of the equation.

$2$ $y$ $-27$ $=$ $-19$

$2$ $y$ $-27$ $+27$ $=$ $-19$ $+27$

$2$ $y$ $=$ $8$ $-27+27$ cancels out

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

$2$ $y$ $=$ $8$

$2$ $y$ $divide2$ $=$ $8$ $divide2$

$y$ $=$ $4$ $2divide2$ cancels out

Check our work

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

$7y-27$ $=$ $5y-19$

$7(4)-27$ $=$ $5(4)-19$ Substitute $y=4$

$28-27$ $=$ $20-19$

$1$ $=$ $1$

Since the equation is true, the answer is correct.

$y=4$
Question 3 of 4

3. Question
Find $y$
- $y=$ (40)
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Inverse Operations

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

Form an equation knowing that the sum of all interior angles of a triangle is $180°$ .

Sum of Interior Angles $=180°$

Angle $1=70°$

Angle $2=(2y-10)°$

Angle $3=y°$

Angle $1$ $+$ Angle $2$ $+$ Angle $3$ $=$ Sum of Interior Angles

$70°$ $+$ $(2y-10)°$ $+$ $y°$ $=$ $180$ Substitute the values

$3y+60$ $=$ $180$

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

Start by moving $60$ to the other side by subtracting $60$ from both sides of the equation.

$3$ $y$ $+60$ $=$ $180$

$3$ $y$ $+60$ $-60$ $=$ $180$ $-60$

$3$ $y$ $=$ $120$ $60-60$ cancels out

Finally, remove $3$ by dividing both sides of the equation by $3$ .

$3$ $y$ $=$ $120$

$3$ $y$ $divide3$ $=$ $120$ $divide3$

$y$ $=$ $40$ $3divide3$ cancels out

Check our work

To confirm our answer, substitute $y=40$ to the formed equation.

$70+(2y-10)+y$ $=$ $180$

$70+(2(40)-10)+40$ $=$ $180$ Substitute $y=40$

$70+(80-10)+40$ $=$ $180$

$70+70+40$ $=$ $180$

$180$ $=$ $180$

Since the equation is true, the answer is correct.

$y=40$
Question 4 of 4

4. Question
Find $x$
- $x=$ (36)
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Inverse Operations

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

Form an equation knowing that the sum of all interior angles of a quadrilateral is $360°$ .

Sum of Interior Angles $=360°$

Angle $1=x°$

Angle $2=4x°$

Angle $3=3x°$

Angle $4=2x°$

Angle $1$ $+$ Angle $2$ $+$ Angle $3$ $+$ Angle $4$ $=$ Sum of Interior Angles

$x$ $+$ $4x$ $+$ $3x$ $+$ $2x$ $=$ $360$ Substitute the values

$10x$ $=$ $360$

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

Start by removing $10$ by dividing both sides of the equation by $10$ .

$10$ $x$ $=$ $360$

$10$ $x$ $divide10$ $=$ $360$ $divide10$

$x$ $=$ $36$ $10divide10$ cancels out

Check our work

To confirm our answer, substitute $x=36$ to the formed equation.

$x+4x+3x+2x$ $=$ $360$

$x+4x+3x+2x$ $=$ $360$ Substitute $x=36$

$36+4(36)+3(36)+2(36)$ $=$ $360$

$36+144+108+72$ $=$ $360$

$360$ $=$ $360$

Since the equation is true, the answer is correct.

$x=36$

Equation Problems (Geometry) 2

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Quiz summary

Information

1. Question

Hint

Good Job!

Inverse Operations

2. Question

Hint

Fantastic!

Inverse Operations

3. Question

Hint

Excellent!

Inverse Operations

4. Question

Hint

Great Work!

Inverse Operations

Quizzes

$3$ $y$ $+15$	$=$	$2$ $y$ $+40$
$3$ $y$ $+15$ $-2y$	$=$	$2$ $y$ $+40$ $-2y$
$y$ $+15$	$=$	$40$	$2y-2y$ cancels out

$y$ $+15$	$=$	$40$
$y$ $+15$ $-15$	$=$	$40$ $-15$
$y$	$=$	$25$	$15-15$ cancels out

$3y+15$	$=$	$2y+40$
$3(25)+15$	$=$	$2(25)+40$	Substitute $y=25$
$75+15$	$=$	$50+40$
$90$	$=$	$90$

$7$ $y$ $-27$	$=$	$5$ $y$ $-19$
$7$ $y$ $-27$ $-5y$	$=$	$5$ $y$ $-19$ $-5y$
$2$ $y$ $-27$	$=$	$-19$	$5y-5y$ cancels out

$2$ $y$ $-27$	$=$	$-19$
$2$ $y$ $-27$ $+27$	$=$	$-19$ $+27$
$2$ $y$	$=$	$8$	$-27+27$ cancels out

$2$ $y$	$=$	$8$
$2$ $y$ $divide2$	$=$	$8$ $divide2$
$y$	$=$	$4$	$2divide2$ cancels out

$7y-27$	$=$	$5y-19$
$7(4)-27$	$=$	$5(4)-19$	Substitute $y=4$
$28-27$	$=$	$20-19$
$1$	$=$	$1$

Angle $1$ $+$ Angle $2$ $+$ Angle $3$	$=$	Sum of Interior Angles
$70°$ $+$ $(2y-10)°$ $+$ $y°$	$=$	$180$	Substitute the values
$3y+60$	$=$	$180$

$3$ $y$ $+60$	$=$	$180$
$3$ $y$ $+60$ $-60$	$=$	$180$ $-60$
$3$ $y$	$=$	$120$	$60-60$ cancels out

$3$ $y$	$=$	$120$
$3$ $y$ $divide3$	$=$	$120$ $divide3$
$y$	$=$	$40$	$3divide3$ cancels out

$10$ $x$	$=$	$360$
$10$ $x$ $divide10$	$=$	$360$ $divide10$
$x$	$=$	$36$	$10divide10$ cancels out

$x+4x+3x+2x$	$=$	$360$
$x+4x+3x+2x$	$=$	$360$	Substitute $x=36$
$36+4(36)+3(36)+2(36)$	$=$	$360$
$36+144+108+72$	$=$	$360$
$360$	$=$	$360$

$70+(2y-10)+y$	$=$	$180$
$70+(2(40)-10)+40$	$=$	$180$	Substitute $y=40$
$70+(80-10)+40$	$=$	$180$
$70+70+40$	$=$	$180$
$180$	$=$	$180$

Angle $1$ $+$ Angle $2$ $+$ Angle $3$ $+$ Angle $4$	$=$	Sum of Interior Angles
$x$ $+$ $4x$ $+$ $3x$ $+$ $2x$	$=$	$360$	Substitute the values
$10x$	$=$	$360$