Solve these simultaneous equations: 3xy = 1, and y = 12x + 3
From first equation: 3xy = 1 => x = 1/(3y)Substitute expression for x into second equation: y = 12x + 3 => y = 12(1/3y) + 3 = 4/y +3Multiply through by y: y2 = 4 + 3y => y2 - 3y - 4 = 0Factorise: (y-4)(y+1) = 0 => y = 4, y = -1 are solutionsx = 1/3y = 1/12, -1/3
Solutions are (1/12,4) and (-1/3, -1)
HM
Answered by Hallam M. • Further Mathematics tutor
3163 Views
See similar Further Mathematics GCSE tutorsRelated Further Mathematics GCSE answers
All answers ▸Why is it that when 'transformation A' is followed by 'transformation B', that the combined transformation is BA and not AB?
Point A lies on the curve y=3x^2+5x+2. The x-coordinate of A is 2. Find the equation of the tangent to the curve at the point A
Show that (n^2) + (n+1)^2 + (n+2)^2 = 3n^2 + 6n + 5, Hence show that the sum of 3 consecutive square numbers is always 2 away from a multiple of 3.
A straight line passes trough the points A(-4;7); B(6;-5); C(8;t). Use an algebraic method to work out the value of t.