Question

Solve for u, z, y, and t:

A;
y/a-b=y/b-a, if a≠b (/ means fractions)
B;
t+ b^2/a=bt/a +a, if a≠b

Answer 1

Where are y and t?

Explanation:

Answer 2

`A.\\\\(y)/(a)-b=(y)/(b)-a\qquad\text{multiply both sides by }\ ab\\eq0\\\\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=(ab^2-a^2b)/(b-a)}\to y=(ab(b-a))/(b-a)\to\boxed{y=ab}`

`B.\\t+(b^2)/(a)=(bt)/(a)+a\qquad\text{multiply both sides by}\ a\\eq0\\\\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)\\eq0\\\\\boxed{t=(a^2-b^2)/(a-b)}\to t=((a-b)(a+b))/(a-b)\to\boxed{t=a+b}`