Question

5x+4y-4=0 and x-12y-20=0 solve in substitution method

Answer 1

Step-by-step explanation:

5x+4y=4and x-2y=20

x=20+2y

5(20+2y)+4y=4

100+10y+4y=4

50+5y+2y=2

50+7y=2

7y=-48

y=-48/7

x=20+2(-48/7)

x= 20-96/7

Answer 2

Answer:

The values of x and y from the given equations by substitution method are 2,$-\frac{3}{2}$ respectively.

The solution is (2,$-\frac{3}{2}$)

Step-by-step explanation:

Given equations are $5x+4y-4=0\hfill (1)$ and

$x-12y-20=0\hfill (2)$

To solve the given equations by substitution method :

From equation (2)  $x-12y-20=0$

$x=12y+20$

• Now substitute the value of x in the equation (1) we have

$5(12y+20)+4y-4=0$

$5(12y)+5(20)+4y-4=0$

$60y+100+4y-4=0$

$64y+96=0$

$64y=-96$

$y=-\frac{96}{64}$

$y=-\frac{3}{2}$

• Substituting the value of y in the equation (2)

$x-12(-\frac{3}{2})-20=0$

$x+18-20=0$

$x=20-18$

$x=2$

Therefore the values of x and y are 2,$-\frac{3}{2}$ respectively.

The solution is (2,$-\frac{3}{2}$)