Question

Solve the following systems using the elimination method.

4x + 7y = 19
2x - y = -13

Answer 1

Answer: x = -4, y = 5

Explanation:

=> 4x + 7y = 19 (Eq. 1)
=> 2x - y = -13 (Eq. 2)

We can multiply Eq. 2 by 7 in order to make the coefficients of ‘y’ same in both the equations thus being able to eliminate ‘y’ and find ‘x’

Therefore,
(2x - y = -13) * 7
=> 14x - 7y = -91 (Eq. 3)

On Adding Eq. 1 and 3, we get :-

18x = -72 (‘y’ gets Eliminated)
x = -72/18
=> x = -4

Therefore, x = -4

Substitute value of ‘x’ in Eq. 2 :-

2x - y = -13
2(-4) - y = -13
-8 - y = -13
-y = -13 + 8
-y = -5
=> y = 5

Therefore, y = 5

Answer 2

Answer: x = -4, y = 5

Explanation:

`$$Solve the following system: \\$\left\{\begin{array}{ll}7 y+4 x=19 & \text { (equation 1) } \\ -y+2 x=-13\end{array} \quad\right.$ (equation 2)\\\\Subtract $(1)/(2) *($ equation 1) from equation 2 :\\$\left\{\begin{array}{l}4 x+7 y=19 \ \ \ \ (\text{equation 1}) \\ 0 x-(9 y)/(2)=-(45)/(2)\end{array}\right.$ (equation 2)\\\\Multiply equation 2 by $-(2)/(9)$ :\\`

`\begin{cases}4 x+7 y=19 & \text { (equation 1) } \\ 0 x+y=5 & \text { (equation 2) }\end{cases}$\\\\Subtract $7 *($ equation 2) from equation 1:\\$\begin{cases}4 x+0 y=-16 & \text { (equation 1) } \\ 0 x+y=5 & \text { (equation 2)\\ }\end{cases}$\\\\Divide equation 1 by 4 :\\$\begin{cases}x+0 y=-4 & \text { (equation 1) } \\ 0 x+y=5 & \text { (equation 2) }\end{cases}$\\\\Answer:\\$$\left\{\begin{array}{l}x=-4 \\y=5\end{array}\right.$$`