Question

if $\triangle def=\triangle pqr$, and $de = 3x - 10$, $qr = 4x - 23$, $pq = 2x + 7$, and $ef = y$, find $x$ and $y$.
$x =$
$y =$

Explanation:

Step1: Find the value of \(x\)

Since \(\triangle DEF=\triangle PQR\), then \(DE = PQ\).
\(3x - 10=2x + 7\)
Subtract \(2x\) from both sides: \(3x-2x-10=2x - 2x+7\), so \(x-10 = 7\).
Add \(10\) to both sides: \(x-10 + 10=7 + 10\), so \(x = 17\).

Step2: Find the value of \(y\)

Since \(\triangle DEF=\triangle PQR\), then \(EF = QR\).
Substitute \(x = 17\) into \(QR = 4x-23\).
\(QR=4\times17-23\)
\(QR = 68 - 23\)
\(QR=45\), so \(y = 45\)

Answer:

\(x = 17\)
\(y = 45\)