Solve the following simultaneous equations: x^2 + y^2 = 12, x - 2y = 3
This is an example of quadratic simultaneous equations. We need to work out the value(s) of x OR y by rearranging one of the equations and then substituting it into the other equation. Once obtaining the x/y value, we have to substitute this value into one of the equations to work out the value of the other (e.g. x if we worked out y first).First, we have to rearrange the linear equation (the one with no x/y squared) to get x or y as the subject. For example, x = 2y + 3. Next, we substitute this equation into x^2 to get "(2y + 3)^2 + y^2 = 12".Now we can solve this equation to work out y. We do this by first expanding, simplifying the expression and then working out y by factorising if we can, using the formula or completing the square. So, we go from "(2y + 3)^2 + y^2 = 12" to "5y^2 + 12y - 3 = 0". As we cannot factorise this, we have to use the formula. Using this method, we get two values for y: 0.23 and -2.63.Finally, we substitute each y value into one of the original equations separately. It would be easiest to use the linear equation, so when substituting y = 0.23 into the equation, we get an x value of 3.456 and when substituting y = -2.63, we get an x value of -2.256.
Related Maths GCSE answers
All answers ▸The rectangles A and B have perimeters of 94cm and 56cm as shown below (insert diagram). Rectangle A: base = 2x cm, height = 3y cm. Rectangle B: base = (x+6)cm, height = (y+4)cm. Use an algebraic method to calculate the area of each rectangle. (8 marks)
Factorise x^2+7x+10 to find the roots of the equatino x^2+7x+10=0
Solve the following definite integral from x=2 to x=-1: ((x^4) + 3(x^2) + 2) dx
The line l is a tangent to the circle x^2 + y^2 = 40 at the point A. A is the point (2,6). The line l crosses the x-axis at the point P. Work out the area of the triangle OAP.