Question

mathematical connections find the value of $n$ that makes $\triangle def \sim \triangle xyz$ when $de = 4$, $ef = 5$, $xy = 4(n + 1)$, $yz = 7n - 1$, and $\angle e \cong \angle y$.
$n = \square$

Explanation:

Step1: Set up proportion for similarity

Since $\triangle DEF \sim \triangle XYZ$ and $\angle E \cong \angle Y$, the sides around the congruent angles are proportional:
$\frac{DE}{XY} = \frac{EF}{YZ}$

Step2: Substitute given values

Replace $DE=4$, $EF=5$, $XY=4(n+1)$, $YZ=7n-1$:
$\frac{4}{4(n+1)} = \frac{5}{7n-1}$

Step3: Simplify left fraction

Cancel 4 in numerator/denominator:
$\frac{1}{n+1} = \frac{5}{7n-1}$

Step4: Cross-multiply to solve for n

Set cross products equal:
$1 \times (7n-1) = 5 \times (n+1)$
$7n - 1 = 5n + 5$

Step5: Isolate n variable

Subtract $5n$ and add 1 to both sides:
$7n - 5n = 5 + 1$
$2n = 6$
$\frac{2n}{2} = \frac{6}{2}$

Answer:

$n=3$