SOLUTION: if the two given equations are being solved so that one unknown is eleminated by the method of comparison what step would be essential? 3x+2y=7 and 6x-5y=8

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: if the two given equations are being solved so that one unknown is eleminated by the method of comparison what step would be essential? 3x+2y=7 and 6x-5y=8      Log On


   



Question 400009: if the two given equations are being solved so that one unknown is eleminated by the method of comparison what step would be essential?
3x+2y=7 and 6x-5y=8

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
3x%2B2y=7 and 6x-5y=8
you can solve this system by SUBSTITUTION or by Elimination/Addition

I will use Elimination/Addition:

Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

3%2Ax%2B2%2Ay=7
6%2Ax-5%2Ay=8

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 6 to some equal number, we could try to get them to the LCM.

Since the LCM of 3 and 6 is 6, we need to multiply both sides of the top equation by 2 and multiply both sides of the bottom equation by -1 like this:

2%2A%283%2Ax%2B2%2Ay%29=%287%29%2A2 Multiply the top equation (both sides) by 2
-1%2A%286%2Ax-5%2Ay%29=%288%29%2A-1 Multiply the bottom equation (both sides) by -1


So after multiplying we get this:
6%2Ax%2B4%2Ay=14
-6%2Ax%2B5%2Ay=-8

Notice how 6 and -6 add to zero (ie 6%2B-6=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%286%2Ax-6%2Ax%29%2B%284%2Ay%2B5%2Ay%29=14-8

%286-6%29%2Ax%2B%284%2B5%29y=14-8

cross%286%2B-6%29%2Ax%2B%284%2B5%29%2Ay=14-8 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:

9%2Ay=6

y=6%2F9 Divide both sides by 9 to solve for y



y=2%2F3 Reduce


Now plug this answer into the top equation 3%2Ax%2B2%2Ay=7 to solve for x

3%2Ax%2B2%282%2F3%29=7 Plug in y=2%2F3


3%2Ax%2B4%2F3=7 Multiply



3%2Ax%2B4%2F3=7 Reduce



3%2Ax=7-4%2F3 Subtract 4%2F3 from both sides

3%2Ax=21%2F3-4%2F3 Make 7 into a fraction with a denominator of 3

3%2Ax=17%2F3 Combine the terms on the right side

cross%28%281%2F3%29%283%29%29%2Ax=%2817%2F3%29%281%2F3%29 Multiply both sides by 1%2F3. This will cancel out 3 on the left side.


x=17%2F9 Multiply the terms on the right side


So our answer is

x=17%2F9, y=2%2F3

which also looks like

(17%2F9, 2%2F3)

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

3%2Ax%2B2%2Ay=7
6%2Ax-5%2Ay=8

we get



graph of 3%2Ax%2B2%2Ay=7 (red) 6%2Ax-5%2Ay=8 (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 (17%2F9,2%2F3). This verifies our answer.