Solve the simultaneous equations 6x - 27 = 15 and 4x + 3y = -3.

Initially we have two unknown variables, so we want to eliminate one of the variables (x) to solve for the other (y). The LCM of 6 and 4 is 12, so multiply each equation such that the coefficient of x is 12:2*(6x - 2y = 15) --> 12x - 4y = 303*(4x + 3y = -3) --> 12x + 9y = -9Subtracting equation 2 from equation 1 eliminates the variable x and gives y = -3. This value is then substituted back into one of the original equations to find the value of x. For example:6x - 2(-3) = 156x = 9x = 1.5Therefore, we have our two solutions whereby x = 1.5, y = -3.

AC
Answered by Alannah C. Maths tutor

3780 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

solve the simultaneous equation 2y + x = 8 and 1 + y = 2x


A right-angled triangle has perpendicular sides of length 6cm and 8cm, and a hypoteneuse of 2y cm. Find the length of y.


A stationary ball starts rolling down a hill, and after 5s it reaches a speed of 12m/s. From here the ground levels off, and the ball continues at this speed for 3 more seconds. Plot this on a velocity-time diagram.


What is the method to solve an equation of type : ax^2+bx+c = 0 ?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning