Respuesta :
Answer: x = -4, y = 5
Step-by-step explanation:
[tex]$Solve the following system: \\$\left\{\begin{array}{ll}7 y+4 x=19 & \text { (equation 1) } \\ -y+2 x=-13\end{array} \quad\right.$ (equation 2)\\\\Subtract $\frac{1}{2} \times($ equation 1) from equation 2 :\\$\left\{\begin{array}{l}4 x+7 y=19 \ \ \ \ (\text{equation 1}) \\ 0 x-\frac{9 y}{2}=-\frac{45}{2}\end{array}\right.$ (equation 2)\\\\Multiply equation 2 by $-\frac{2}{9}$ :\\[/tex]
[tex]\begin{cases}4 x+7 y=19 & \text { (equation 1) } \\ 0 x+y=5 & \text { (equation 2) }\end{cases}$\\\\Subtract $7 \times($ 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.$[/tex]
Answer: x = -4, y = 5
Step-by-Step 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
Step-by-Step 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