# Thread: 5q-2(q+3)=3(q-2)

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2. ## Re: 5q-2(q+3)=3(q-2)

$\displaystyle 5q-2(q+3)=3(q-2)$

Distribute the $\displaystyle -2$ and $\displaystyle 3$ to expand the respective brackets, then collect like terms, and it becomes a case of simple manipulation to finish. Please show how much you can do.

Edit: I assume you are trying to show that both sides are equal. Otherwise, there's nothing you can really do here.

3. ## Re: 5q-2(q+3)=3(q-2)

Expand brackets, then group like terms, what do you get?

4. ## Re: 5q-2(q+3)=3(q-2)

5q-2(q+3) = 3(q-2)

5q - 2q - 6 = 3q - 6

3q - 6 = 3q - 6

3q - 3q = 6 - 6

0 = 0

5. ## Re: 5q-2(q+3)=3(q-2) Originally Posted by mathland 5q-2(q+3) = 3(q-2)

5q - 2q - 6 = 3q - 6

3q - 6 = 3q - 6

3q - 3q = 6 - 6

0 = 0

There is no solution in terms of this equation. Both sides of the equation cancel out.
Actually there are infinitely many solutions to this equation, because ANY value for q will satisfy the equation.

6. ## Re: 5q-2(q+3)=3(q-2) Originally Posted by Prove It Actually there are infinitely many solutions to this equation, because ANY value for q will satisfy the equation.
Thanks for correcting me. Yes, you are right. On this site and others like it, we learn from each other.

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