Solve the simultaneous equations: 1) 4x - 2y = 28, 2) 4y - 3x = -36.

Before starting, look whether either equation has common factors. Equation 1 does common factor 2, so divide it by 2 both sides to get 1) 2x - y = 14. Rearrange 1) to get one unknown in terms of the other. In this case, y as follows: 2x = 14 + y, 2x -14 = y. Substitute y in the second equation 2) using y = 2x - 14: 4(2x - 14) -3x = -36,going to 8x - 56 -3x = -36, move numbers to right and simplify x terms on the left: 8x - 3x = 56 - 36 giving 5x = 20 dividing by 5 to get x = 4. Now x = 4 is known, so substitute into the first equation 1) y = 2x - 14 to get y: y = 2(4) -14 going to y = 8 - 14, hence y = -6. Final Answer: x = 4, y = -6

Answered by Aaron H. Maths tutor

3653 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Write 7/8 as a decimal


Solve the simultaneous equations: y=x/2 + 2 and 2y+3x=12


Show that (4+√12)(5-√3)= 14+6√3


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.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences