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Two equations are shown: • Equation 1: 3 4 (x−12)=12 •Equation 2: 3 4 y−12=12 Solve each equation. Then, enter a number in each box to make this statement true. The value of x is: The value of y is:

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Answer

[tex]x=28[/tex]

[tex]y=32[/tex]

Explanation

Let's solve each equation step by step

Equation 1: [tex]\frac{3}{4} (x-12)=12[/tex]

Step 1. Distribute [tex]\frac{3}{4}[/tex]

[tex]\frac{3}{4} x-\frac{3}{4} *12=12[/tex]

[tex]\frac{3}{4} x-9=12[/tex]

Step 2. Add 9 to both sides

[tex]\frac{3}{4} x-9+9=12+9[/tex]

[tex]\frac{3}{4} x=21[/tex]

Step 3. Multiply both sides of the equation by [tex]\frac{4}{3}[/tex]

[tex]\frac{3}{4} x*\frac{4}{3}=21*\frac{4}{3}[/tex]

[tex]x=28[/tex]

Equation 2: [tex]\frac{3}{4} y-12=12[/tex]

Step 1. Add 12 to both sides

[tex]\frac{3}{4} y-12+12=12+12[/tex]

[tex]\frac{3}{4} y=24[/tex]

Step 2. Multiply both sides by [tex]\frac{4}{3}[/tex]

[tex]\frac{3}{4} y*\frac{4}{3}= 24*\frac{4}{3}[/tex]

[tex]y=32[/tex]

Answer:

The required value of x = 28

The required value of y = 32

Step-by-step explanation:

The given two equations are :

[tex]\frac{3}{4}(x-12)=12..........(1)\\\\\frac{3}{4}\cdot y-12=12...........(2)[/tex]

Now, we need to solve both the equations and find out the required values of both the unknowns x and y.

Solving equation (1) to obtain value of x. We have,

[tex]\frac{3}{4}(x-12)=12\\\\\text{On cross multiplication, we get}\\\\\implies 3\times (x-12)=12\times 4\\\implies 3x -36=48\\\implies 3x=48+36\\\implies 3x=84\\\bf\implies x=28[/tex]

Solving equation (2) to obtain value of y. We have,

[tex]\frac{3}{4}\cdot y-12=12\\\\\implies \frac{3}{4}\cdot y=12+12\\\\\implies \frac{3}{4}\cdot y=24\\\\\text{On cross multiplicatrion, We get}\\\implies 3\cdot y=24\times 4\\\implies 3y = 96\\\bf\implies y=32[/tex]

Hence, The required value of x = 28

And The value of y = 32

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