Simultaneously solve these equations 3x+y=7 and 3x-y=5

Method 1 (Elimination) -

You can see that If you add the 2 equations together you can eliminate the y variable like so 6x=12, then if you divide both sides by 6 you get x=2. Then if you place x=2 back into either of the equations you get y=1.

Method 2 (Substitution) -

Take equation 1 and rearrange it so you get y in terms of x, so all the y's on one side and all the x's on the other side. You get y=7-3x. Take this expression for y and put it into equation 2, 3x-y=5. You get 3x - (7-3x) = 5. If you expand out the brackets you get 6x-7=5 and so 6x=12 then dividing both sides by 6, x=2. Like before place x=2 into either equation to get y=1.

CB

Related Maths GCSE answers

All answers ▸

Factorise 4x+6x^2


Why do the denominators have to be equal when adding fractions, but not when multiplying them?


Factorise 2c2 + 8c + 8.


How do you break down a wordy question (e.g. Aled has three concrete slabs. Two slabs square, of length x, & the third rectangular of dimensions 1m & x+1m. Show 2x^2 +x-6=0 & Solve this)