Question

based on the measures shown, could the figure be a parallelogram?
yes, one pair of opposite sides could measure 10 in., and the other pair could measure 13 in.
yes, one pair of opposite sides could measure 10 in., and the other pair could measure 8 in.
no, there are three different values for x when each expression is set equal to 10.
no, the value of x that makes one pair of sides congruent does not make the other pair of sides congruent.

Explanation:

Step1: Solve for \(x\) when \(x + 5=2x - 3\)

$$
LATEXBLOCK0
$$

Step2: Substitute \(x = 8\) into \(x + 2\)

\(x+2=8 + 2=10\)

Step3: Check if both pairs of opposite sides are equal

When \(x = 8\), \(x + 5=8+5 = 13\) and \(2x-3=2\times8-3=13\), \(x + 2=10\) (one side is \(10\)). But for a parallelogram, both pairs of opposite sides must be equal. If we assume \(10\) is one side, solving \(x+5=x + 2\) gives \(5=2\) (impossible). Solving \(10=2x - 3\) gives \(x=\frac{13}{2}\), then \(x + 5=\frac{13}{2}+5=\frac{23}{2}\), \(x + 2=\frac{13}{2}+2=\frac{17}{2}\). Solving \(10=x + 5\) gives \(x = 5\), then \(2x-3=2\times5-3 = 7\), \(x + 2=5+2=7\). But when \(x = 5\), \(2x-3=7
eq10\). The value of \(x\) that makes one pair of sides congruent (\(x = 5\) for \(x + 2=2x - 3\)) does not make the other pair (\(10\) and \(x + 5\)) congruent.

Answer:

No, the value of \(x\) that makes one pair of sides congruent does not make the other pair of sides congruent.