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I'm currently working on some homework, and I'm very stuck on how to solve an algebraic equation. It should be simple for me, but it's just not coming back. Any help?

$1224$ = $(6)^2h+\frac{(6)^2(42-h)}{3}$

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$$1224 = (6)^2 h + \frac {(6)^2 (42-h)}{3}$$

Change $(6)^2$ to $36$ and give the terms common denominators for a start.

$$1224 = \frac {3(36h)}{3} + \frac {36(42 - h)}{3}$$

Multiply both sides by $3$.

$$3672 = 108h + 1512 - 36h$$

Subtract $1512$ from both sides and combine the $xh$ terms.

$$2160 = 72h$$

Divide by $72$ to compute $h$.

$$h = 30$$

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Write $$1224=(6^2)h+\frac{6^2(42)}{3} + \frac{-(6)^2}{3}h$$ then collect the $h$'s together and move the constants to the other side.

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$1224 = \frac{3(6^2)h}{3} + \frac{(6^2)(42 - h)}{3}$

$1224 = \frac{3(36)h}{3} + \frac{(36)(42 - h)}{3}$

$1224 = \frac{3(36)h+(36)(42 - h)}{3}$

$3672 = 3(36)h+(36)(42 - h)$

$3672 = 108h + 1512 - 36h$

$3672 = 72h + 1512$

$2160 = 72h$

$30 = h$

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$1224=36h+36(42-h)/3$

$1224=36h+(1512-36h/3)$

$1224=36h+504-12h$

$1224=24h+504$

$720=24h$

$30=h, \qquad\text{so }h\text{ is }30$.

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    $\begingroup$ Unfortunately, the formatting on this answer makes it difficult to read. In the future, I would encourage you to at least use spaces between lines of equation. Further, you may be interested in looking at latex or mathjax for how to improve the appearance of math $\endgroup$ – davidlowryduda May 6 '13 at 23:12

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