How do you solve $3 (g - 1) - 7 = 3 g + 4$ $3(g-1)-7=3g+4$ ?

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Answer 1 · 2018-04-24T12:29:11

There is no solution.

Solve:

$3 (g - 1) - 7 = 3 g + 4$ $3(g-1)-7=3g+4$

Distribute the $3$ $3$ across $(g - 1)$ $(g-1)$ using the distributive property: $a (b + c) = a b + a c$ $a(b+c)=ab+ac$ .

$3 g - 3 - 7 = 3 g + 4$ $3g-3-7=3g+4$

Simplify.

$3 g - 10 = 3 g + 4$ $3g-10=3g+4$

Cancel $3 g$ $3g$ on both sides.

$- 10 = 4$ $-10=4$

Since $- 10 \neq 4$ $-10!=4$ , there is no solution.

How do you solve 3(g-1)-7=3g+43(g−1)−7=3g+43(g-1)-7=3g+4?