How do solve the following linear system?: # x-2y=8 , 4x=-11+y #?

Question

How do solve the following linear system?: #  x-2y=8 ,  4x=-11+y    #?

Eli Assouline · Accepted Answer

I found:
#x=-30/7#
#y=-43/7# We can isolate #x# from the first equation and substitute into the second to find #y#: #x=8+2y# So: #4(8+2y)=-11+y# #32+8y=-11+y# #7y=-43# #y=-43/7# We use this value back into the first equation: #x=8+2(-43/7)=(56-86)/7=-30/7#

Savannah Anderson · Answer

undefined To solve the linear system $x - 2y = 8$ and $4x = -11 + y$, we can use the substitution method or the elimination method. Let's use the elimination method:

First, we rearrange the equations to put them in standard form:

$x - 2y = 8$   (Equation 1)

$4x - y = -11$  (Equation 2)

Now, let's eliminate one variable. We can eliminate the variable $y$:

Multiply Equation 1 by 4 to match the coefficient of $y$ in Equation 2:

$4(x - 2y) = 4(8)$

$4x - 8y = 32$

Now, subtract Equation 2 from this new equation:

$4x - 8y - (4x - y) = 32 - (-11)$

$4x - 8y - 4x + y = 32 + 11$

$-7y = 43$

Now, solve for $y$:

$y = \frac{43}{-7}$

$y = -\frac{43}{7}$

Now that we have found the value of $y$, we can substitute it back into one of the original equations to solve for $x$. Let's use Equation 1:

$x - 2(-\frac{43}{7}) = 8$

$x + \frac{86}{7} = 8$

Subtract $\frac{86}{7}$ from both sides:

$x = 8 - \frac{86}{7}$

Now, find a common denominator:

$x = \frac{56}{7} - \frac{86}{7}$

$x = \frac{-30}{7}$

So, the solution to the linear system is $x = \frac{-30}{7}$ and $y = -\frac{43}{7}$.