Question

what is the value of x in the equation $\frac{2}{3}x - 7 = 5 - \frac{3}{5}x$?
a $3\frac{18}{19}$
b $9\frac{9}{19}$
c $15\frac{1}{5}$
d $19\frac{1}{5}$

Explanation:

Step1: Add $\frac{3}{5}x$ and 7 to both sides

To solve for \( x \), we first want to get all the \( x \)-terms on one side and the constants on the other. So we add \(\frac{3}{5}x\) to both sides and add 7 to both sides of the equation \(\frac{2}{3}x - 7 = 5 - \frac{3}{5}x\).

$$ \frac{2}{3}x + \frac{3}{5}x - 7 + 7 = 5 + 7 - \frac{3}{5}x + \frac{3}{5}x $$

Simplifying the left side, the \(-7\) and \(+7\) cancel out, and on the right side, the \(-\frac{3}{5}x\) and \(+\frac{3}{5}x\) cancel out. So we have:

$$ \frac{2}{3}x + \frac{3}{5}x = 12 $$

Step2: Combine the \( x \)-terms

To combine \(\frac{2}{3}x\) and \(\frac{3}{5}x\), we need a common denominator for the fractions. The least common denominator of 3 and 5 is 15. So we rewrite the fractions with denominator 15:

$$ \frac{2}{3}x = \frac{2 \times 5}{3 \times 5}x = \frac{10}{15}x $$

$$ \frac{3}{5}x = \frac{3 \times 3}{5 \times 3}x = \frac{9}{15}x $$

Now we can add the two fractions:

$$ \frac{10}{15}x + \frac{9}{15}x = \frac{10 + 9}{15}x = \frac{19}{15}x $$

So our equation becomes:

$$ \frac{19}{15}x = 12 $$

Step3: Solve for \( x \)

To solve for \( x \), we multiply both sides of the equation by the reciprocal of \(\frac{19}{15}\), which is \(\frac{15}{19}\):

$$ x = 12 \times \frac{15}{19} $$

Calculating the right side:

$$ x = \frac{12 \times 15}{19} = \frac{180}{19} $$

Now we convert \(\frac{180}{19}\) to a mixed number. We divide 180 by 19: \( 19 \times 9 = 171 \), and \( 180 - 171 = 9 \). So:

$$ x = 9\frac{9}{19} $$

Answer:

B. \( 9\frac{9}{19} \)