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find a if xz = 68, xy = 12a - 4 and yz = 3a - 3.

Question

find a if xz = 68, xy = 12a - 4 and yz = 3a - 3.

Explanation:

Step1: Use segment addition postulate

According to the segment addition postulate, \(XY + YZ=XZ\). Substitute the given expressions: \((12a - 4)+(3a - 3)=68\).

Step2: Simplify the left - hand side

Combine like terms: \(12a+3a-4 - 3=68\), which simplifies to \(15a-7 = 68\).

Step3: Solve for \(a\)

Add \(7\) to both sides: \(15a-7 + 7=68 + 7\), so \(15a=75\). Then divide both sides by \(15\): \(a=\frac{75}{15}\).

Answer:

\(a = 5\)