# Solve the simultaneous equations y = 2x-3 and x^2 +y^2 = 2

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Let the first equation be equation 1 and the second equation be equation 2.

Firstly, you must substitute equation 1 into equation 2. This is because there are currently two unknown values in each equation (x and y) and therefore you must eliminate one of them so that you just have x's or y's in a singe equation.

Therefore, lets sub in equation 1 into equation 2. But first we must square:

From equation 1: y2 = (2x-3)2 = (2x-3)(2x-3) = 4x2 -12x +9

Then substituting this into equation 2:  x2 +y2  = 2

x2 +4x2-12x+9 =2

5x2 -12x+9 =2

Subtracting 2 from both sides of the equation:

5x2 -12x + 7 = 0

(5x -7)(x-1) = 0

Now to solve for x: either of the brackets need to equal 0 so that the left hand side of the equation equals 0.

Therefore, 5x-7 = 0, which with some simple rearrangement gives rise to x = 7/5

x-1 = 0, which with some simple rearrangement gives rise to x = 1

Please note we have two values for x; this is because we have a quadratic equation (one which has powers of 2 in it). As we have two values for x it means we have two corresponding values for y. To find these values substitute the values of x you have just found back into equation 1:

From equation 1: When x = 7/5, y = -1/5

When x = 1, y = -1