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

3691 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

2476 people are at a football match. The ratio of men to women is 3 : 1. How many more men than women are at the match?


Work out the points of intersection of the graphs of y= (x+2)(x-4) and y=3x+6.


What is completing the square and how do I do it?


v^2 = 2w - x^2. w = 40; x = 4. Find the value of v.


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:

© 2025 by IXL Learning