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Algebra/Changing Coefficients in both equations



I know I need to multiply and subtract somewhere, but I need a step by step process on how to find the solution.

There are several approaches, but here is one that works the best.
First, get a 1 as the x-coefficient in 1st equation by dividing all terms in that equation by 2.  When the term is divided by 2, I put it at the end.  This is to note that 3y/2 is really 3y/2.  If we put 3/2y, it could be read as 3/(2y), and that is incorrect.  Since the equation is 2x + 3y = 9, dividing each term by 2 gives x + 3y/2 = 9/2.

Since the x-coefficient in the 2nd equation is 3, add -3 times the new 1st equation to the 2nd equation.  Multiplying the new 1st equation by -3 gives -3x - 9y/2 = -27/2.

When we try to add this to 3x + 2y = 11, we need to have the 2y and the 11 in havles.
Multiplying each of these terms by 2/2 gives 3x + 4y/2 = 22/2.

The answer is then 3x - 3x -9y/2 + 4y/2 = -27/2 + 22/2 => -5y/2 = -5/2.
Multiplying both sides of this equation by -2/5 gives y = 1.

Looking at the equations at the top, if y = 1, we have 2x + 3 = 9 and 3x + 2 = 11.
Adding -3 to both sides on the 1st equation gives 2x = 6, so x = 3.
Checking this by adding -2 to both sides on the 2nd equation gives 3x = 9, so x = 3.

Just to make sure this is right, put x = 3 and y = 1 back in.
For the 1st equation, this gives 2*3 + 3 = 6 + 3 = 9, and that is correct.
For the 2nd equation,  this gives 3*3 + 2 = 9 + 2 = 11, and that is correct.

From this, the answer is correct.


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Scott A Wilson


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