MATHS

Asked by Caitlin GraceYou should rearrange the second equation for either x or y, both ways will work. If y=2-x, then change all the y's in the first equation to 2-x and simplify. You should have a quadratic which when you solve will give you two x values. Using those you can work out the corresponding y values using x+y=2

The squared equation is a circle who’s centre is the origin and radius 3 units. The linear equation is a straight line of Y=-x + 2 Solo of -1 and crossing y axis at 2 and so crossing the x axis at 2 also. It would help you understand the question if you sketch it. Solving it you substitute the linear equation for y into the circle equation X^2 + (2-x)^2 =9 So X^2+4-4x + x^2 =9 Giving 2x^2 -4x - 5 =0 Use solution of quadratic equation X= ((4/4)+or - (Root(16+40))/4) X= 2.87 or -0.87

X^2+y^2=9 Equation A X+y=2 Equation B Make y subject of Equation B Y=2-x Therefore y^2=(2-x)^2 Substitute y^2 for (2-x)^2 in Equation A Therefore: X^2+(2-x)^2=9 Expanding (2-x)^2: (2-x)^2= (2-x)(2-x) =4-4x+x^2 Therefore X^2+X^2-4x+4=9 Simplifying this: 2x^2-4x+4=9 Now solve for X: Use formula for quadratic equations: X= ((-b) +/- (b^2-4ac)^1/2)/2a With A=2, B=(-4), C=4 Substitute the vales you find for x into Equation B with y as the subject (y=2-x) to find the corresponding values of y that satisfy both equations. Round answers to 3 significant figures if needed to make working out easier.