Solve by elimination with multiplication

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Monsterscore

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C1620B1D-8852-4002-B270-F5234F36CD87.jpegI need some help with solving by elimination with multiplication I’m confused on where to mulit
 
What have you tried? Please show YOUR work.

Hint: What is the Least Common Multiple of 2 and 3?
 
The coefficient of the x's are 2 and 3. You want these numbers to be the same. You need to multiply the 1st equation by any number you choose (except 0) and multiply the 2nd equation by any number you choose but in the end the coefficient of the x's must be the same.
Give this a try.

BTW, you can do this same procedure to get the coefficients of the y's to be the same.
 
Jomo said:
The coefficient of the x's are 2 and 3. You want these numbers to be the same. You need to multiply the 1st equation by any number you choose (except 0) and multiply the 2nd equation by any number you choose but in the end the coefficient of the x's must be the same.
Give this a try.

BTW, you can do this same procedure to get the coefficients of the y's to be the same.
Click to expand...
Don’t you have to multiply that number to the whole entire equation?
 
Yes, you need to multiply the whole equation by this number. Otherwise it will be like just crossing out the x's from the two equations.
 
A lot of beginners in mathematics get the idea that there exist one correct method to solve a mathematics problem. That's seldom true. What you need to do is read a problem carefully and use your knowledge of basic arithmetic to decide how to solve the problem.

Here, you have the two equations
2x+ 3y= 9 and
3x+ 4y= 5.

I know how to solve a single equation in a single unknown (and I was told to solve this "using elimination") so I want to combine those two equations in to one "eliminating" one of the unknowns.

Which one? Either one! It's our choice.

I can decide to eliminate x. I see that I have "2x" in one equation and "3x" in the other. The "least common multiple" of 2 and 3 is 6. I can get 6x in the first equation by multiplying 2x by 3. Of course I have to multiply the entire equation to do that: 6x+ 9y= 27. I can also get 6x in the second equation by multiplying the whole equation by 2: 6x+ 8y= 10.

So now we have 6x+ 9y= 27 and 6x+ 8y= 10. Subtracting the second equation from the first "eliminates" x and gives y= 17. Replace y by 17 in either of the original equations to get an equation for x.

OR

Choose to eliminate y first. The first equation has "3y" and the second has "4y". The least common multiple is 12y. Multiplying the first equation by 4 gives 8x+ 12y= 36. Multiplying the second equation by 3 9x+ 12y= 15. Subtracting the first equation from the second "eliminates" y and leaves x= -21. You can replace x in either of the original equations to find y. If you hold your tongue just right it will be the same "y= 17" we got before!
 
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