Problem 28
Question
Solve each equation. $$ \frac{3}{2}(y+4)=\frac{20-y}{2} $$
Step-by-Step Solution
Verified Answer
The solution is \(y = 2\).
1Step 1: Eliminate the Fractions
To make the equation easier to work with, eliminate fractions by multiplying every term by 2, which is the common denominator. This gives: \[2 \times \frac{3}{2}(y+4) = 2 \times \frac{20-y}{2}\]Simplifying, we get:\[3(y+4) = 20-y\]
2Step 2: Distribute and Simplify
Apply the distributive property to remove the parentheses:\[3y + 12 = 20 - y\]
3Step 3: Combine Like Terms
Move all terms involving the variable \(y\) to one side of the equation and constant terms to the other:\[3y + y = 20 - 12\]This simplifies to:\[4y = 8\]
4Step 4: Solve for y
Divide each side by 4 to solve for \(y\):\[y = \frac{8}{4}\]Thus, \(y = 2\).
Key Concepts
Eliminating FractionsDistributive PropertyCombining Like Terms
Eliminating Fractions
When you encounter fractions in an equation, a helpful strategy is to eliminate them. This makes the equation simpler and more straightforward to solve. The core idea is to multiply every term in the equation by the least common denominator. In the original equation \(\frac{3}{2}(y+4)=\frac{20-y}{2}\), the denominator is 2. By multiplying every term by 2, you clear the fractions:
\[2 \times \frac{3}{2}(y+4) = 2 \times \frac{20-y}{2}\]
The fractions are canceled out because the multiplication by 2 and division by 2 neutralize each other. You're left with:
\[3(y+4) = 20-y\]
This equation is much easier to work with. It sets the stage for further solving steps, reducing the complexity as the fractions disappear.
\[2 \times \frac{3}{2}(y+4) = 2 \times \frac{20-y}{2}\]
The fractions are canceled out because the multiplication by 2 and division by 2 neutralize each other. You're left with:
\[3(y+4) = 20-y\]
This equation is much easier to work with. It sets the stage for further solving steps, reducing the complexity as the fractions disappear.
Distributive Property
The distributive property is a key algebraic principle that helps in simplifying equations with parentheses. It states that \(a(b+c) = ab + ac\). This means you distribute the multiplication over the addition inside the parentheses.
In our example, you need to apply this to the term \(3(y+4)\). Let's see how that works:
\[3(y+4) = 3 \times y + 3 \times 4\]
After distributing, you will simplify the expression to:
\[3y + 12\]
This simplification process helps in clarifying the equation, removing the parentheses and helping you see what terms you have to work with. It’s a crucial step in solving linear equations efficiently.
In our example, you need to apply this to the term \(3(y+4)\). Let's see how that works:
\[3(y+4) = 3 \times y + 3 \times 4\]
After distributing, you will simplify the expression to:
\[3y + 12\]
This simplification process helps in clarifying the equation, removing the parentheses and helping you see what terms you have to work with. It’s a crucial step in solving linear equations efficiently.
Combining Like Terms
Once you have applied the distributive property, the next step is to combine like terms. Like terms are terms that involve the same variable to the same power. In the equation derived from our example, you have:
\[3y + 12 = 20 - y\]
To make solving straightforward, you want all the \(y\) terms on one side of the equation. Start by moving \(-y\) to the left side by adding \(y\) to both sides:
\[3y + y + 12 = 20\]
This combines like terms on the left:
\[4y + 12 = 20\]
Then, keep only the terms involving \(y\) on one side by subtracting 12 from both sides:
\[4y = 8\]
This process simplifies the equation, isolating the variable on one side, and it is a crucial step to solving for that variable. Combining like terms reduces the complexity and helps in easily finding the solution for \(y\).
\[3y + 12 = 20 - y\]
To make solving straightforward, you want all the \(y\) terms on one side of the equation. Start by moving \(-y\) to the left side by adding \(y\) to both sides:
\[3y + y + 12 = 20\]
This combines like terms on the left:
\[4y + 12 = 20\]
Then, keep only the terms involving \(y\) on one side by subtracting 12 from both sides:
\[4y = 8\]
This process simplifies the equation, isolating the variable on one side, and it is a crucial step to solving for that variable. Combining like terms reduces the complexity and helps in easily finding the solution for \(y\).
Other exercises in this chapter
Problem 28
Solve each equation. \(|4 x-24.8|=32.4\)
View solution Problem 28
Solve each compound inequality, if possible. Graph the solution set (if one exists) and write it using interval notation. $$ 5 x-3 \geq 2 \text { and } 6 \geq 4
View solution Problem 29
Determine whether the relation defines \(y\) to be a function of \(x .\) If it does not, find two ordered pairs where more than one value of \(y\) corresponds t
View solution Problem 29
Factor each polynomial. $$ 23 a^{2} b^{3}+4 x^{3} y^{2} $$
View solution