Drag each step and justification to the correct location on the table. Each step and justification can be used more than once, but not all steps and justifications will be used.
Order each step and justification that is needed to solve the equation below:
[tex]$\frac{2}{3} y+15=9$[/tex]
Multiplication property of equality [tex]$\frac{2}{3} y \cdot \frac{3}{2}=6 \cdot \frac{3}{2}$[/tex] [tex]$\frac{2}{3} y=-6$[/tex]
Subtraction property of equality [tex]$y=-9$[/tex] [tex]$\frac{2}{3} y=6$[/tex] [tex]$y=9$[/tex]
[tex]$\frac{2}{3} y \cdot \frac{3}{2}=-6 \cdot \frac{3}{2}$[/tex]
Steps
Justifications
[tex]$\frac{2}{3} y+15=9$[/tex]
Given
[tex]$\frac{2}{3} y+15-15=9-15$[/tex]
Simplification
Simplification
Answer (1)
Subtract 15 from both sides: 3 2 y + 15 − 15 = 9 − 15 .
Simplify: 3 2 y = − 6 .
Multiply both sides by 2 3 : 3 2 y ⋅ 2 3 = − 6 ⋅ 2 3 .
Simplify to find the value of y: y = − 9 .
Explanation
Problem Analysis We are given the equation 3 2 y + 15 = 9 and we need to solve for y . We will do this by isolating y on one side of the equation.
Subtracting 15 from both sides First, we subtract 15 from both sides of the equation to isolate the term with y : 3 2 y + 15 − 15 = 9 − 15
Simplification Simplifying both sides, we get: 3 2 y = − 6
Multiplying by the reciprocal Next, we multiply both sides of the equation by 2 3 to isolate y : 3 2 y ⋅ 2 3 = − 6 ⋅ 2 3
Simplification Simplifying both sides, we find the value of y : y = − 9
Examples
Imagine you're baking a cake and need to adjust the recipe. If 3 2 of a cup of sugar plus 15 grams of vanilla is too sweet and you want it to equal 9 grams of sweetness, this equation helps you find out how much sugar you need to remove. Solving linear equations like this is essential for adjusting quantities in recipes, managing budgets, or calculating distances and speeds. By understanding how to isolate variables, you can easily adapt formulas to fit your specific needs and achieve the desired outcome.
