How do you solve the following linear system: #4x + 5y = -3 , -x + y = 3#?

Question

Replaced a #+# with #=# for the first equation; hopefully, this is what was intended.

Avery Alexander · Accepted Answer

#x=-2#, #y=-1#.

From the second equation, we know that #y=3+x#. Thus, the initial equation becomes #4x+5color(green)(y)=-3 -> 4x+5color(green)((3+x))=-3# Expand the left memeber: #4x+15+5x = 9x+15# So, #9x+15=-3 \iff 9x = -18 iff x=-2# Knowing this, we obtain #y#, since #y=3+x=3-2=1#

Ethan Adkins · Answer

#(x,y)=(-2,1)# Given [1]#color(white)("XXX")4x+5y=-3# [2]#color(white)("XXX")-x+y=3# Multiply [2] by #4# [3]#color(white)("XXX")-4x+4y=12# Add [1] and [3] [4]#color(white)("XXX")9y=9# Divide [4] by #9# [5]#color(white)("XXX")y=1# Substitute #1# for #y# in [2] [6]#color(white)("XXX")-x+(1)=3# Subtract #1# from both sides of [6] [7]#color(white)("XXX")-x=2# Multiply both sides of [7] by #(-1)# [8]#color(white)("XXX")x=-2#

Jameson Allen · Answer

undefined To solve the linear system 4x + 5y = -3 and -x + y = 3, you can use the method of substitution or elimination.

Let's use the method of elimination:

Multiply the second equation by 4 to match the coefficients of x:

-4x + 4y = 12

Now, add this equation to the first equation:

(4x + 5y) + (-4x + 4y) = -3 + 12

Combine like terms:

9y = 9

Divide both sides by 9:

y = 1

Now, substitute y = 1 into either of the original equations. Let's use the second equation:

-x + 1 = 3

Add x to both sides:

x = -2

So, the solution to the linear system is x = -2 and y = 1.