Solve these simultaneous equations: 2x + y = 7, and 3x - y = 8. Do so by 1) Eliminating an Unknown and 2) Substitution.

  1. The number 'in front of' an unknown is called the Coefficient. When eliminating an unknown, look to see whether the coefficients of x are the same in both equations, or if the coefficients of y are the same. In this case, the y coefficients are both 1, and so we can 'get rid of' y from our equations.

Adding the equations together we get; 2x + 3x + y - y = 7 + 8  --->  5x + 0 = 15  --->   x = 3           The y's cancel out.

Then subsitute x = 3 into either equation, we'll use the first;   2(3) + y = 7  --->  6 + y = 7  --->  y = 1 

  1. By substitution. Rearrange one of the equations to get x = ... or y = ... . If we rearrange the first equation we get  y = 7 - 2x.

Now we can substitute  y = 7 - 2x into the second equation;  3x - (7 - 2x) = 8. Use brackets here so that we don't get confused with signs.

Expand this out;  3x - 7 + 2x = 8  --->  5x = 15  --->  x = 3

Then substitute x = 3 into our 'y' equation;   y = 7 - 2(3)  --->  y = 7 - 6   --->   y = 1 

We get x = 3 and y = 1 with both methods.

EH
Answered by Ellie H. Maths tutor

3584 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Anne picks a 4-digit number. The first digit is not zero. The 4-digit number is a multiple of 5. How many different 4-digit numbers could she pick?


If 3x+6x+3=21, find the value of x


Examine the sequence 1 + 1/3 + (1/3)^2 + … + (1/3)^n Under what condition does it converge? What does it converge to?


At what points does the line y = x +1 intersect the circle x^2 + y^2 + 18x + 20y + 81 = 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