Change the equation to show what it would look like if you were solving for \( b \).
Show steps, please.
[tex]A= \frac{1}{2}h(b+c) [/tex]
Answer (3)
A = 2 1 h ( b + c ) ∣ m u lt i pl y b y 2 2 a = h ( b + c ) ∣ d i v i d e b y h h 2 a = b + c ∣ s u b t r a c t c h 2 a − c = b S o l u t i o n i s b = h 2 a − c
[A= \frac{1}{2}h(b+c) \ 2A=h(b+c)\ 2A=hb+hc\ hb=2A-hc\ b=\frac{2A-hc}{h}
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To solve for b in A = 2 1 h ( b + c ) , multiply by 2 then divide by h to isolate b + c . Finally, subtract c from both sides to get b = h 2 A − c . This gives you b in terms of A , h , and c .
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