How do you solve $(7t + 7)/4 + (7t + 8)/3 = 33$ ?

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Answer 1 · 2015-11-29T02:59:09

$t=7$

Multiply both sides by $12$ because $4$ and $3$ are factors of $12$

$12((7t+7)/4+(7t+8)/3)=33(12)$

$3(7t+7)+4(7t+8)=396$

Distribute the $3$ and $4$

$21t+21+28t+32=396$

Combine like terms

$49t+53=396$

Subtract $53$ from both sides

$49t=343$

Divide both sides by $49$

$t=7$

Now check to see if $t=7$ gives $33$

$(7(7)+7)/4+(7(7)+8)/3=33$

$(49+7)/4+(49+8)/3=33$

$56/4+57/3=33$

$14+19=33$

How do you solve (7t + 7)/4 + (7t + 8)/3 = 33(7t + 7)/4 + (7t + 8)/3 = 33?