How do you solve the following system?: 4x3y=5,x2y=15

1 Answer
May 17, 2018

x=5andy=5

Explanation:

4x3y=5eqn1

x2y=15eqn2

From the eqn1 rearrange the equation..

4x3y=5

Multiplying through by minus();

(4x3y=5)

4x+3y=5eqn1

Now solving simultaneously..

4x+3y=5eqn1

x2y=15eqn2

Using Elimination Method..

Multiplying eqn1 by 1 and eqn2 by 4 respectively..

1(4x+3y=5)

4(x2y=15)

4x+3y=5eqn3

4x8y=60eqn4

Subtracting eqn4 from eqn3

(4x4x)+(3y(8y)=5(60)

0+3y+8y=5+60

11y=55

y=5511

y=5

Substituting the value of y into eqn2

x2y=15eqn2

x2(5)=15

x10=15

x=15+10

x=5

Therefore;

x=5andy=5