How do I solve simultaneous equations like 2x + 5y = 50 and 3x + y = 23?

With simultaneous equations like these, you first want to get to a point where you have one equation with only one variable. To do this, you must eliminate one of the variables. In this case, if you multiply both sides of the second equation by 5, you get 15x + 5y = 115. Both equations now have a 5y term in them, so you can take one away from the other and eliminate the variable y:
15x + 5y = 115-(2x + 5y = 50)
13x + 0y = 65
65 / 13 = 5, so x = 5. We can now plug this value back into either of the original equations to find y. Using the second equation (before we multiplied it by 5), we get:
3 * 5 + y = 2315 + y = 23y = 8
So x = 5 and y = 8

AH

Related Maths 13 Plus answers

All answers ▸

Factorise fully the following: x^2 + 7x + 10.


Solve (1/2)*(4x+2)+3=8


The distance of a particle from its start point with respect to time is given by the equation s = 5t^2 + 2t + 4, what is the acceleration of the particle at time t?


Bob has £500 and gives 70% to Edmund and 40% of what is left to George. Work out how much money each person has.