How do solve the following linear system?: #-4x + 4 = -4y, 7x + 6y + 11 = 0 #?

Question

Savannah Anderson · Accepted Answer

The solution is #(-5/13, 8/13)# This problem makes things fairly easy for you by providing a clear definition of #y# in the first equation: #-4x + 4 = -4y# #x - 1 = y# We can plug this into the other equation to solve for #x# and then use it to find #y#: #7x + 6(x - 1) + 11 = 0# #7x + 6x - 6 + 11 = 0# #13x + 5 = 0# #x = -5/13# Then: #-5/13 + 1 = y# #y = 8/13# So the solution is #(-5/13, 8/13)# A little additional info on the problem in concluding. In case it is not clear, the solution takes the form of a point, #(x,y)#. This is because solving a system of equations is the same as finding the point(s) at which those lines (or curves) intersect.

Eli Assouline · Answer

undefined To solve the linear system:

Equation 1: $-4x + 4 = -4y$

Equation 2: $7x + 6y + 11 = 0$

Rearrange Equation 1 to solve for $y$:

$y = -x + 1$

Substitute this expression for $y$ into Equation 2:

$7x + 6(-x + 1) + 11 = 0$

Now, solve for $x$:

$7x - 6x + 6 + 11 = 0$

$x + 17 = 0$

$x = -17$

Now, substitute $x = -17$ into the expression for $y$:

$y = -(-17) + 1$

$y = 17 + 1$

$y = 18$

So, the solution to the linear system is $x = -17$ and $y = 18$.