Question

2x-3y=5 and7x+4y=9 by substitution method

Answer 1

Given equations

$2x-3y=5$

$2x=5+3y$

$x=\dfrac{5+3y}{2}$

Substituting the value of x in the second equation, we get,

$7x+4y=9$

$\implies 7(\dfrac{5+3y}{2} ) +4y=9$

$\implies \dfrac{35+21y}{2} +4y=9$

$\implies 35+21y+8y=18$

$\implies 29y=18-35$

$\implies \boxed{\boxed{\bold{ y = \dfrac{-17}{29} }}}}}}}}$

Substituting the value of y in x, we get,

$\implies \boxed{\boxed{\bold{x=\frac{30}{29} }}}}}$

                                                   

Answer 2

To find:

Isolate the x and y.

Solutions:

$\displaystyle \mathsf{2x-3y=5}}$

Add by 3y from both sides of an equation.

$\displaystyle \mathsf{2x-3y+3y=5+3y}}$

Solve.

$\displaystyle \mathsf{2x=5+3y}}$

Then, divide by 2 from both sides of an equation.

$\displaystyle \mathsf{\frac{2x}{2}=\frac{5}{2}+\frac{3y}{2} }}$

Solve.

$\displaystyle \mathsf{x=\frac{5+3y}{2}}}$

Substitution on x=5+3y/2.

$\displaystyle \mathsf{7*\frac{5+3y}{2}+4y=9 }}$

Use distributive property. (expand.)

$\displaystyle \mathsf{DISTRIBUTIVE \ PROPERTY}}$

$\displaystyle \mathsf{A(B+C)=AB+AC}}$

$\displaystyle \mathsf{7*\frac{5+3y}{2}+4y=9=y=-\frac{17}{29} }}$

Solve.

$\displaystyle \mathsf{\frac{5+3\left(\displaystyle -\frac{17}{29}\right)}{2}=\frac{47}{29}}}$

Subtract.

47-17=30

$\Large\boxed{\mathsf{y=-\frac{17}{29}, x=\frac{30}{29} }}}$

Answer:

So, the solutions is y=-17/29, and x=30/29.