Solve the simultaneous equations algebraically.

x^2 + y^2 = 29y - x = 3Rearrange the second equation such that one variable, either x or y, is the subject. I will rearrange to make x the subject. x = y - 3. Substitute the new equation into equation 1 in place of x.(y-3)^2 + y^2 = 29.Rearrange to make the equation equal 0 and simplify. y^2 - 6y + 9 + y^2 = 29. 2y^2 - 6y - 20 = 0. All variables have a factor of 2 and so we can simplify further by dividing by 2.y^2 - 3y - 10 = 0. Factorise. (y-5)(y+2)=0. Therefore, y=5 when x=? and y=-2 when x=?. To find out what x is in each case of y we can substitute the x values back into one of the original equations and see what values of y we receive. In this case I will substitute back into equation 2. When y=5, 5 - x = 3. Therefore, x=2. --> When y=5, x=2.When y=-2, -2 - x= 3. Therefore, x = -5. --> When y=-2, x=-5.

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Answered by Jane A. Maths tutor

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