Question7 Complete the equation so that it has infinitely many solutions. $4\left(\frac{x}{6}-2\right)=\frac{1}{4}(2 x-$ $\qquad$

Studdy Solution

STEP 1

What is this asking? We need to find the missing number that makes both sides of the equation equal, no matter what value $x$ takes. Watch out! Don't forget to distribute correctly when multiplying through parentheses!

STEP 2

1. Simplify the left side.
2. Simplify the right side.
3. Match the sides.

STEP 3

Let's distribute the 4 on the left side of the equation: $4 \cdot \frac{x}{6} - 4 \cdot 2$ This simplifies to: $\frac{4x}{6} - 8$

STEP 4

We can simplify the fraction $\frac{4x}{6}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 2: $\frac{4x \div 2}{6 \div 2} = \frac{2x}{3}$ So the left side of the equation becomes: $\frac{2x}{3} - 8$

STEP 5

Let's distribute $\frac{1}{4}$ on the right side.
We'll represent the missing number with the variable $c$ : $\frac{1}{4} \cdot 2x - \frac{1}{4} \cdot c$ This simplifies to: $\frac{2x}{4} - \frac{c}{4}$

STEP 6

We can simplify the fraction $\frac{2x}{4}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 2: $\frac{2x \div 2}{4 \div 2} = \frac{x}{2}$ So the right side becomes: $\frac{x}{2} - \frac{c}{4}$

STEP 7

To have infinitely many solutions, the $x$ terms on both sides must be equal.
We need to make $\frac{2x}{3}$ equal to $\frac{x}{2}$ .
We can do this by multiplying both the numerator and denominator of $\frac{x}{2}$ by $\frac{2}{3}$ : $\frac{x \cdot \frac{2}{3}}{2 \cdot \frac{2}{3}} = \frac{\frac{2x}{3}}{\frac{4}{3}} = \frac{2x}{3} \cdot \frac{3}{4} = \frac{2x \cdot 3}{3 \cdot 4} = \frac{6x}{12} = \frac{x \cdot 6}{2 \cdot 6} = \frac{x}{2}$ However, since the problem wants us to have infinitely many solutions, we need the $x$ terms on both sides to be equal, and they already are!

STEP 8

Now, let's make the constant terms match.
We have $-8$ on the left side and $-\frac{c}{4}$ on the right side.
We set them equal to each other: $-8 = -\frac{c}{4}$ To solve for $c$ , we multiply both sides by $-4$ : $-8 \cdot -4 = -\frac{c}{4} \cdot -4$ $32 = c$ So, the missing number is 32.

STEP 9

The completed equation is: $4\left(\frac{x}{6}-2\right)=\frac{1}{4}(2 x-32)$

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