How would you solve the system of these two linear equations: #2x + 3y = -1# and #x - 2y = 3#? Enter your solution as an ordered pair (x,y).

Question

Sadie Allen · Accepted Answer

#(1, -1)#

Given,

#{(2x + 3y = -1 color(white)(xxxxxx) "---(I)"), (x - 2y = 3 color(white)(xxxxxxxxx) "---(II)") :}#

Multiply #"(II)"# by #2# and then subtract from #"(I)"#,

#7y = -7# #color(white)(x)y = -1#

Putting #y = -1# in #"(I)"#,

#2x + 3(-1) = -1#

#color(white)(xxxxxxx) 2x = -1 + 3#

#color(white)(xxxxxxx) 2x = 2#

#color(white)(xxxxxxxx) x = 1#.

Hence, #x = 1# and #y = -1#

Allison Allen · Answer

undefined To solve the system of linear equations 2x + 3y = -1 and x - 2y = 3, you can use the method of substitution or elimination. Here, we'll use the method of substitution.

From the second equation x - 2y = 3, we can rearrange it to get x = 3 + 2y.

Now substitute this expression for x into the first equation: 2(3 + 2y) + 3y = -1.

Solve for y: 6 + 4y + 3y = -1 => 7y = -7 => y = -1.

Now that we have the value of y, substitute it back into either equation to find the value of x. Let's use the second equation x - 2y = 3: x - 2(-1) = 3 => x + 2 = 3 => x = 1.

Therefore, the solution to the system of equations is (1, -1).