(b) Solve the following equations $$ \frac{4}{2 y-5}-\frac{3}{y+3}=0 $$. 6 (a) Factorise these quadratic expression

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Solve the equation $$ \frac{4}{2y-5}-\frac{3}{y+3}=0 $$.
Solve the equation by following steps:
- step0: Solve for $$y$$:
  $$\frac{4}{2y-5}-\frac{3}{y+3}=0$$
- step1: Find the domain:
  $$\frac{4}{2y-5}-\frac{3}{y+3}=0,y \in \left(-\infty,-3ight)\cup \left(-3,\frac{5}{2}ight)\cup \left(\frac{5}{2},+\inftyight)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{4}{2y-5}-\frac{3}{y+3}ight)\left(2y-5ight)\left(y+3ight)=0	imes \left(2y-5ight)\left(y+3ight)$$
- step3: Simplify the equation:
  $$-2y+27=0$$
- step4: Move the constant to the right side:
  $$-2y=0-27$$
- step5: Remove 0:
  $$-2y=-27$$
- step6: Change the signs:
  $$2y=27$$
- step7: Divide both sides:
  $$\frac{2y}{2}=\frac{27}{2}$$
- step8: Divide the numbers:
  $$y=\frac{27}{2}$$
- step9: Check if the solution is in the defined range:
  $$y=\frac{27}{2},y \in \left(-\infty,-3ight)\cup \left(-3,\frac{5}{2}ight)\cup \left(\frac{5}{2},+\inftyight)$$
- step10: Find the intersection:
  $$y=\frac{27}{2}$$
The solution to the equation $$ \frac{4}{2y-5}-\frac{3}{y+3}=0 $$ is $$ y = \frac{27}{2} $$ or $$ y = 13.5 $$.
$$ y = \frac{27}{2} $$ or $$ y = 13.5 $$.

(b) Solve the following equations .
6 (a) Factorise these quadratic expression

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(b) Solve the following equations . 6 (a) Factorise these quadratic expression