Question 31802
Hello!
In order to do elimination using multiplication, the idea is to multiply one of the equations by a given number so that the coefficients of either x or y are the same in both equations. Let me illustrate:

You have:
{{{2x+6y=10}}}
{{{5x+3y=1}}}

Look at the coefficients of x. In the 1st equation it's 2; in the second one it's 5. You could multiply the first equation by 2.5 and you would get another 5 as the x coefficient in it. The first equation would become:

{{{2x+6y=10}}}
{{{2.5(2x+6y)=2.5(10)}}}
{{{5x + 15y = 25}}}

And the nyou would be ready to eliminate. Just subtract the 2nd equation from the 1st one, getting:

{{{ (5x - 5x) + (15y - 3y) = 25 - 1}}}
{{{ 0 + 12y = 24}}}
{{{12y = 24}}}

And now it's easy to find the value of y; and afterwards, of x.


In the example I showed you, I did a multiplication such that the x would 'disappear' after the subtraction. Of course, you could have done the same with y instead of x. The original system was:

{{{system(2x+6y=10,5x+3y=1)}}}

Look at the coefficients of y: 6 in the 1st equation and 3 in the 2nd one. If you multiplied the 2nd equation by 2, you would get a '6y' in it. Let's do it:

{{{2*(5x+3y)=2*1}}}
{{{10x+6y = 2}}}

And then you could subtract the 1st equation from this one, getting:

{{{(10x-2x) + (6y - 6y) = 2 - 10}}}
{{{8x + 0 = -8}}}
{{{8x = -8}}}

So now we can easily find x, and then y by simple subtitution.


I hope this helps!

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