We need to get rid of the fractions so we cross-multiply both equations by the common denominator (x+y)(x-y). This expression when expanded is x^2-y^2. The two equations become: 5(x-y)-2(x+y)=-x^2+y^2 and 15(x-y)+7(x+y)=10(x^2-y^2). Opening the brackets and adding the x and y terms we get: 3x-7y=-x^2+y^2 and 22x-8y=10(x^2-y^2). The squared terms need to be removed because they complicate the problem. The first equation has -(x^2-y^2) and the second has 10(x^2-y^2). We can substitute -3x+7y, which is the same as 7y-3x, from the first equation in the second equation so that we eliminate the squared terms: the second equation becomes 22x-8y=10(7y-3x)=70y-30x. From this we get a relationship between x and y: 52x=78y, so x=3y/2. We can now substitute this value for x in 3x-7y=-x^2+y^2, so we get: 9y/2-7y=-9y^2/4+y^2. Multiply the whole equation by 4 to get rid of the fractions: 18y-28y=-9y^2+4y^2. -10y=-5y^2. We can divide both sides by -5y to get 2=y or y=2. Now we can find x from x=3y/2=3. To check the answer substitute the values of x and y in the original equations.