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Chapter 4: Problem 64

For Problems 55-70, solve each equation for the indicated variable. (Objective 4) $$ \frac{x+1}{3}=\frac{y-5}{2} \text { for } y $$

Short Answer

Expert verified
\( y = \frac{2x + 17}{3} \)

Step by step solution

01

Cross-Multiply to Eliminate Fractions

Start with the equation \(\frac{x+1}{3} = \frac{y-5}{2}\). To eliminate the fractions, cross-multiply, resulting in \(2(x+1) = 3(y-5)\).
02

Distribute Both Sides

Distribute the 2 on the left side and the 3 on the right side: \(2x + 2 = 3y - 15\).
03

Isolate Terms Involving y

Add 15 to both sides to move the constant term: \(2x + 17 = 3y\).
04

Solve for y

Divide both sides by 3 to solve for \(y\): \(y = \frac{2x + 17}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
Solving equations is like finding the answer to a puzzle. We want to find the value of a variable that makes the equation true. In our example, we need to solve for the variable \(y\) in the equation \(\frac{x+1}{3} = \frac{y-5}{2}\). An equation is like a balance scale. Whatever we do to one side, we must do to the other, so it remains equal. When solving equations:
  • Identify the variable you need to solve for. In our case, it's \(y\).
  • Perform the same operation on both sides of the equation to maintain balance.
  • Continue until the variable is by itself on one side of the equation.
Remember, equations are solved gradually, by using mathematical operations, and one step, maintaining balance, leads to another until the variable is isolated.
Cross-Multiplication
Cross-multiplication is a technique used to solve equations involving fractions. When you have an equation like \(\frac{x+1}{3} = \frac{y-5}{2}\), you can use cross-multiplication to get rid of the fractions.Here's how it works:
  • Multiply the numerator of the first fraction by the denominator of the second fraction. This gives you \(2(x+1)\).
  • Multiply the numerator of the second fraction by the denominator of the first fraction. This gives you \(3(y-5)\).
By cross-multiplying, the equation becomes: \[ 2(x+1) = 3(y-5) \]This method helps simplify equations by removing fractions, making it easier to isolate the variable in later steps.
Variable Isolation
Variable isolation involves manipulating an equation so that the variable of interest is alone on one side. It's like unwrapping a gift to find what's inside. After distributing, we have the equation \(2x + 2 = 3y - 15\).To isolate \(y\):
  • Add 15 to each side to undo the subtraction of 15 from \(3y\). This balances the equation, resulting in \(2x + 17 = 3y\).
  • Finally, divide each side by 3 to separate \(y\) from its coefficient. This gives us \(y = \frac{2x + 17}{3}\).
Through these steps, we successfully isolate \(y\), showing its relationship to \(x\). The whole process involves balancing moves that help "unwrap" the variable from its surrounding terms.
Distributive Property
The distributive property is a fundamental algebraic rule that helps simplify expressions. It states that multiplying a sum by a number gives the same result as multiplying each addend separately and then adding the products. In our example, we distributed the numbers 2 and 3 within the equation:Starting with:\[ 2(x+1) = 3(y-5) \]We apply the distributive property to both sides:
  • Left side: \(2 \times x + 2 \times 1 = 2x + 2\)
  • Right side: \(3 \times y - 3 \times 5 = 3y - 15\)
So, the equation becomes \(2x + 2 = 3y - 15\). The distributive property helps simplify expressions, being crucial when dealing with equations, as it helps set up for variable isolation and solving. Always apply this property carefully to maintain the balance of the equation.

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Most popular questions from this chapter

For Problems 1-12, solve each equation. You will be using these types of equations in Problems \(13-41\). $$ 0.7(15)-x=0.6(15-x) $$

Two cars start from the same place traveling in opposite directions. One car travels 4 miles per hour faster than the other car. Find their speeds if after 5 hours they are 520 miles apart.

Suppose that Lou invested a certain amount of money at \(3 \%\) interest, and he invested \(\$ 750\) more than that amount at \(5 \%\). His total yearly interest was \(\$ 157.50\). How much did he invest at each rate?

For Problems 55-70, solve each equation for the indicated variable. (Objective 4) $$ 9 x-6 y=13 \quad \text { for } y $$

A sum of \(\$ 2300\) is invested, part of it at \(10 \%\) interest and the remainder at \(12 \%\). If the interest earned by the \(12 \%\) investment is \(\$ 100\) more than the interest earned by the \(10 \%\) investment, find the amount invested at each rate.
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