SOLUTION: Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c. {3x−y=14 4x+3y=−3 (x,

Algebra ->  Inequalities -> SOLUTION: Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c. {3x−y=14 4x+3y=−3 (x,      Log On


   

Question 1131809: Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c.
{3x−y=14
4x+3y=−3
(x, y) =
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) 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

3%2Ax-1%2Ay=14
4%2Ax%2B3%2Ay=-3

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 3 and 4 to some equal number, we could try to get them to the LCM.

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

4%2A%283%2Ax-1%2Ay%29=%2814%29%2A4 Multiply the top equation (both sides) by 4
-3%2A%284%2Ax%2B3%2Ay%29=%28-3%29%2A-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
12%2Ax-4%2Ay=56
-12%2Ax-9%2Ay=9

Notice how 12 and -12 add to zero (ie 12%2B-12=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%2812%2Ax-12%2Ax%29-4%2Ay-9%2Ay%29=56%2B9

%2812-12%29%2Ax-4-9%29y=56%2B9

cross%2812%2B-12%29%2Ax%2B%28-4-9%29%2Ay=56%2B9 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

-13%2Ay=65

y=65%2F-13 Divide both sides by -13 to solve for y



y=-5 Reduce


Now plug this answer into the top equation 3%2Ax-1%2Ay=14 to solve for x

3%2Ax-1%28-5%29=14 Plug in y=-5


3%2Ax%2B5=14 Multiply



3%2Ax=14-5 Subtract 5 from both sides

3%2Ax=9 Combine the terms on the right side

cross%28%281%2F3%29%283%29%29%2Ax=%289%29%281%2F3%29 Multiply both sides by 1%2F3. This will cancel out 3 on the left side.


x=3 Multiply the terms on the right side


So our answer is

x=3, y=-5

which also looks like

(3, -5)

Notice if we graph the equations (if you need help with graphing, check out this solver)

3%2Ax-1%2Ay=14
4%2Ax%2B3%2Ay=-3

we get



graph of 3%2Ax-1%2Ay=14 (red) 4%2Ax%2B3%2Ay=-3 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (3,-5). This verifies our answer.


(x, y) =(3, -5)

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c.
{3x−y=14
4x+3y=−3
(x, y) =
Don't do this the way the woman did it. It's a very INEFFICIENT way to solve this simple problem.

All you need to do is:

1)  Multiply the 1st equation: (3x - y = 14) by 3 to get eq (iii), or 9x - 3y = 42 

2)  Add eqs (iii) & (ii) to eliminate "y" and get the value of "x."

3)  Substitute the value of x into any of the 2 ORIGINAL equations, and solve to get the value of "y."

That's it. Nothing more, nothing less!!