Question

solve for j.
\\(\frac{1}{j + 8} = \frac{7j}{7j}\\)

Explanation:

Step1: Simplify the right - hand side

First, simplify the fraction $\frac{7j}{7j}$. For $j
eq0$, we can cancel out the common factor $7j$ in the numerator and the denominator. So $\frac{7j}{7j} = 1$ (when $j
eq0$). The equation becomes $\frac{1}{j + 8}=1$.

Step2: Solve for j

Multiply both sides of the equation $\frac{1}{j + 8}=1$ by $j + 8$ (assuming $j+8
eq0$, i.e., $j
eq - 8$). We get $1\times1=j + 8$. Then, subtract 8 from both sides of the equation: $j=1 - 8=-7$.
We need to check if this solution is valid. For $j=-7$, the original equation's denominators:

• For the left - hand side denominator $j + 8=-7 + 8 = 1

eq0$.

• For the right - hand side denominator $7j=7\times(-7)=-49

eq0$. Also, when we simplified $\frac{7j}{7j}$, we had the condition $j
eq0$, and $j = - 7
eq0$. So $j=-7$ is a valid solution.

Answer:

$j=-7$