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 $y$ $y$
- $y =$ $y=$ (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.

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

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

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

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

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

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

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

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

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

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

Check our work

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

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

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

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

$90$ $90$ $=$ $=$ $90$ $90$

Since the equation is true, the answer is correct.

$y = 25$ $y=25$
Question 2 of 4

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.

Form an equation knowing that corresponding angles are equal.

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

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

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

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

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

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

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

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

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

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

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

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

$2$ $2$ $y$ $y$ $\div 2$ $divide2$ $=$ $=$ $8$ $8$ $\div 2$ $divide2$

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

Check our work

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

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

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

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

$1$ $1$ $=$ $=$ $1$ $1$

Since the equation is true, the answer is correct.

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

3. Question
Find $y$ $y$
- $y =$ $y=$ (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 $180 °$ $180°$ .

Sum of Interior Angles $= 180 °$ $=180°$

Angle $1 = 70 °$ $1=70°$

Angle $2 = (2 y - 10) °$ $2=(2y-10)°$

Angle $3 = y °$ $3=y°$

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

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

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

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

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

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

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

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

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

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

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

$y$ $=$ $40$ $3divide3$ 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

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

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

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

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

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

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

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

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

$70+(2y-10)+y$	$=$	$180$
$70+(2(40)-10)+40$	$=$	$180$	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
$10x$	$=$	$360$

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

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