$A.\\\\\dfrac{y}{a}-b=\dfrac{y}{b}-a\qquad\text{multiply both sides by }\ ab\neq0\\\\by-ab^2=ay-a^2b\qquad\text{add}\ ab^2\ \text{to both sides}\\\\by=ay+ab^2-a^2b\qquad\text{subtract}\ ay\ \text{from both sides}\\\\by-ay=ab^2-a^2b\qquad\text{distributive}\\\\(b-a)y=ab^2-a^2b\qquad\text{divide both sides by}\ (b-a)\\\\\boxed{y=\dfrac{ab^2-a^2b}{b-a}}\to y=\dfrac{ab(b-a)}{b-a}\to\boxed{y=ab}$

$B.\\t+\dfrac{b^2}{a}=\dfrac{bt}{a}+a\qquad\text{multiply both sides by}\ a\neq0\\\\at+b^2=bt+a^2\qquad\text{subtract}\ b^2\ \text{from both sides}\\\\at=bt+a^2-b^2\qquad\text{subtract}\ bt\ \text{from both sides}\\\\at-bt=a^2-b^2\qquad\text{distributive}\\\\(a-b)t=a^2-b^2\qquad\text{divide both sides by}\ (a-b)\neq0\\\\\boxed{t=\dfrac{a^2-b^2}{a-b}}\to t=\dfrac{(a-b)(a+b)}{a-b}\to\boxed{t=a+b}$

8 0

3 years ago

Can anyone help me with this it’s precalculus

andrew-mc [135]

The x intercepts are where there are points on the x axis. In this case your x intercepts would be (-3,0), (3.5,0) and (5,0) and if it asks for what x equals it’s going to be x=-3, x=3.5 and x=5. Hope this helps

5 0

4 years ago