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How do you solve #4x + y = 26# and #5x - 2y =13#?

Answer 1

Eva Adamson

The solution is the point #(5,6)#.

Equation 1: #4x+y=26#

Equation 2: #5x-2y=13#

We can use addition/elimination and substitution to solve this system.

Multiply Equation 1 by #2#.

#2(4x+y)=26xx2#

Simplify.

#8x+2y=52#

Add Equation 1 and 2.

#8x+2y=52# #5x-2y=13# #-----# #13x##color(white)(.......)=65#

Divide both sides by #13#.

#x=65/13#

#x=5#

Substitute #5# for #x# in Equation 2 and solve for #y#.

#5(5)-2y=13#

#25-2y=13#

Subtract #25# from both sides.

#-2y=13-25#

#-2y=-12#

Divide both sides by #-2#.

#y=(-12)/(-2)# #larr# Two negatives make a positive.

#y=6#

The solution is the point #(5,6)#.

graph{(4x+y-26)(5x-2y-13)=0 [-9.33, 10.67, 0.08, 10.08]}

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Answer 2

Eva Adamson

To solve the system of equations (4x + y = 26) and (5x - 2y = 13), you can use the method of substitution or elimination. Solving by elimination yields (x = 3) and (y = 14).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.