Answered

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Suppose M is the midpoint of line FG.

A) FM= 5y+13, MG=5-3y, FG=?

B) FM=8a+1, FG=42, a=?

Sagot :

[tex]A)\\|FM|=5y+13;\ |MG|=5-3y\\\\|FG|=|FM|+|MG|\\\\|FG|=(5y+13)+(5-3y)=2y+18\\\\B)\\|FM|=8a+1;\ |FG|=42\\\\|FG|=2|FM|\\\\2(8a+1)=42\\16a+2=42\ \ \ \ \ |subtract\ 2\ from\ both\ sides\\16=40\ \ \ \ \ |divide\ both\ sides\ by\ 16\\a=\frac{40}{16}\\a=\frac{5}{2}\\a=2.5[/tex]
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