b) $$ x+y=2 $$ and $$ 2 x+5 y=9 $$ c) $$ \frac{3 x}{2}+\frac{4 y}{5}=-1 $$ and $$ 2 x+\frac{y}{2}=\frac{3}{2} $$

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Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+y=2\2x+5y=9\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=2-y\2x+5y=9\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$2\left(2-yight)+5y=9$$
- step3: Simplify:
  $$4+3y=9$$
- step4: Move the constant to the right side:
  $$3y=9-4$$
- step5: Subtract the numbers:
  $$3y=5$$
- step6: Divide both sides:
  $$\frac{3y}{3}=\frac{5}{3}$$
- step7: Divide the numbers:
  $$y=\frac{5}{3}$$
- step8: Substitute the value of $$y:$$
  $$x=2-\frac{5}{3}$$
- step9: Calculate:
  $$x=\frac{1}{3}$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{1}{3}\y=\frac{5}{3}\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{1}{3}\y=\frac{5}{3}\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(\frac{1}{3},\frac{5}{3}ight)$$
Solve the system of equations $$ \frac{3 x}{2}+\frac{4 y}{5}=-1;2 x+\frac{y}{2}=\frac{3}{2} $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}\frac{3x}{2}+\frac{4y}{5}=-1\2x+\frac{y}{2}=\frac{3}{2}\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}\frac{3x}{2}+\frac{4y}{5}=-1\x=\frac{3-y}{4}\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$\frac{3	imes \frac{3-y}{4}}{2}+\frac{4y}{5}=-1$$
- step3: Simplify:
  $$\frac{3\left(3-yight)}{8}+\frac{4y}{5}=-1$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{3\left(3-yight)}{8}+\frac{4y}{5}ight)	imes 40=-40$$
- step5: Simplify the equation:
  $$45+17y=-40$$
- step6: Move the constant to the right side:
  $$17y=-40-45$$
- step7: Subtract the numbers:
  $$17y=-85$$
- step8: Divide both sides:
  $$\frac{17y}{17}=\frac{-85}{17}$$
- step9: Divide the numbers:
  $$y=-5$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{3-\left(-5ight)}{4}$$
- step11: Substitute back:
  $$x=\frac{3+5}{4}$$
- step12: Calculate:
  $$x=2$$
- step13: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=-5\end{array}ight.$$
- step14: Check the solution:
  $$\left\{ \begin{array}{l}x=2\y=-5\end{array}ight.$$
- step15: Rewrite:
  $$\left(x,yight) = \left(2,-5ight)$$
Let's solve the two systems of equations step by step.

### Part b: 
The equations are:
1. $$ x + y = 2 $$
2. $$ 2x + 5y = 9 $$

Using the solution, we find:
- The solution to this system is $$ (x, y) = \left(\frac{1}{3}, \frac{5}{3}ight) $$ or approximately $$ (0.33, 1.67) $$.

### Part c:
The equations are:
1. $$ \frac{3x}{2} + \frac{4y}{5} = -1 $$
2. $$ 2x + \frac{y}{2} = \frac{3}{2} $$

Using the solution, we find:
- The solution to this system is $$ (x, y) = (2, -5) $$.

### Summary of Solutions:
- For part b: $$ (x, y) = \left(\frac{1}{3}, \frac{5}{3}ight) $$
- For part c: $$ (x, y) = (2, -5) $$
The solutions are: - For part b: $$ x = \frac{1}{3} $$, $$ y = \frac{5}{3} $$ - For part c: $$ x = 2 $$, $$ y = -5 $$

b) \( x+y=2 \) and \( 2 x+5 y=9 \) c) \( \frac{3 x}{2}+\frac{4 y}{5}=-1 \) and \( 2 x+\frac{y}{2}=\frac{3}{2} \)

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