Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 153352This question is from textbook elementary and intermediate algebra concepts and applications
: Can I please get some help solving these
Solve by substitution or elimination method:
-4x + 3y = 5
12x - 9y = -15
This question is from textbook elementary and intermediate algebra concepts and applications
: Can I please get some help solving these
Solve by substitution or elimination method:
-4x + 3y = 5
12x - 9y = -15

Answer by jim_thompson5910(9368) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

-4*x+3*y=5
12*x-9*y=-15

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get -4 and 12 to some equal number, we could try to get them to the LCM.

Since the LCM of -4 and 12 is -12, we need to multiply both sides of the top equation by 3 and multiply both sides of the bottom equation by 1 like this:

3*(-4*x+3*y)=(5)*3 Multiply the top equation (both sides) by 3
1*(12*x-9*y)=(-15)*1 Multiply the bottom equation (both sides) by 1


So after multiplying we get this:
-12*x+9*y=15
12*x-9*y=-15

Notice how -12 and 12 add to zero, 9 and -9 add to zero, 15 and -15 and to zero (ie -12+12=0) 9+-9=0, and 15+-15=0)


So we're left with

0=0


which means any x or y value will satisfy the system of equations. So there are an infinite number of solutions


So this system is dependent