Question

Please need help on 2 math questions

Answer 1

Question 1:

For this case we have the following expression:

$\frac {\frac {y-1} {y ^ 2-36}} {\frac {1-7y} {y + 6}} =\\\frac {(7y-1) (y + 6)} {(y ^ 2-36) (1-7y)} =$

We have to:

$y ^ 2-36 = (y + 6) (y-6)$

Rewriting:

$\frac {(7y-1) (y + 6)} {(y + 6) (y-6) (1-7y)} =\\\frac {7y-1} {(y-6) (1-7y)} =$

We take common factor "-" in the denominator:

$\frac {7y-1} {(y-6) * - (- 1 + 7y)} =\\\frac {7y-1} {- (y-6) * (7y-1)} =\\- \frac {1} {(y-6)}$

ANswer:

$- \frac {1} {(y-6)}$

Question 2:

For this case we must simplify the following expression:

$(3n ^ 4 + 1) + (- 8n ^ 4 + 3) - (- 8n ^ 4 + 2) =$

We eliminate parentheses keeping in mind that:

$+ * - = -\\- * - = +\\3n ^ 4 + 1-8n ^ 4 + 3 + 8n ^ 4-2 =$

We add similar terms:

$3n ^ 4-8n ^ 4 + 8n ^ 4 + 1 + 3-2 =\\3n ^ 4 + 2$

Answer:

$3n ^ 4 + 2$