# What are x and y given that 3x + 2y = 9 and 7x – 5y = 0?

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We have the two equations:

3x + 2y = 9 … (1)

7x – 5y = 0 … (2)

5*(1) + 2*(2)

=> 15x + 10y + 14x – 10y = 45

=> 29x = 45

=> x = 45/29

Substituting this in (1)

3(45/29) + 2y = 9

=> 135/29 + 2y =9

=> 2y = 9- (135 / 29)

=> y = 63/29

**Therefore x = 45/29 and y = 63/29.**

We have the equations:

3x + 2y = 9 ..............(1)

7x - 5y = 0.................(2)

We need to solve for x and y.

We will use the elimination method to solve.

We will multiply (1) by 5 and (2) by 2

==> 5*(1) ==> 15x + 10y = 45

==> 2*(2) ==> 14x - 10y = 0

Now we will add both equations.

==> 29 x = 45

**==> x = 45/29 **

Now we will substitute into (1) to find y.

==> 3x +2y = 9

==> 3(45/29) + 2y = 9

==> 135/29 + 2y = 9

==> 2y = 9 - 135/29 = 126/29

==> y= 126/29*2 = 63/29

**==> y= 63/29**

I'm given two equations:

1) 3x + 2y = 9

2) 7x - 5y = 0

I'm going to use the elimination method and

in order to do so, I must 5*1) + 2*2).

=> {15x + 10y = 45} + {14x - 10y = 0}

=> 29x = 45

Therefore, **x= 45/29**

and if we plug in 45/29 into x in equation 1), than we get

=> 675/29 + 10y = 45

=> 10y = 630/29

Therefore,** y = 63/29**