How do you solve these maths questions?

[MoSalahsBeard]

Can anyone help? Son is stuck on these.

On the first one the difference before the = sign is an extra x and the difference after the sign is 3. So 1x must be 3. You use that knowledge to work out what the rest are.

So 2x must be 6, 6 plus y is 10 so y must be 4.

You can subtract one equation from another.
2x + y = 10
-(x + y = 7)
2x - x = x, y - y = 0, 10 - 7 = 3
x = 3
Then you can substitute 3 for x.
3 + y = 7

It's a question of finding how to combine the equations in such a way as to cancel out one of the unknowns.

For (a) and (b) you can subtract them, but (c) is different: you have a +y and a -y so add them together.

If (2x+y)=14 and (x-y)=1 then (2x+y)+(x-y) = 14+1

in other words 2x + y + x - y = 15
so 2x + x + y - y = 15
so 3x = 15
(don't forget the final step, x=5, y=4)

(d) is similar to (c). (e) is a bit more complicated - you'll need to multiply the statements first. (Same as you do with fractions to get a common denominator)

Has he not been told how to?

Because looking at the second set, those are about graphs, so I'd guess he's been told to draw both lines on the same graph and the point you're looking for is where they intercept.

However much more fun is to do them as simultaneous equations.

Easiest method is:
2x+y =10
x + y = 7

Using 2nd equation: y = 7-x

Put that into equation 1:
2x + (7 - x) = 10
x + 7 = 10
Therefore x = 3, y = 4 (find y by putting x into one of the equations)

Or (harder but more fun, and better method because it can be easier with harder ones:

take first equation and take away second:

(2x +y) - (x +y) = 10-7
2x + y - x- y = 3
x = 3

The reason why that's longer lasting is if you have something like:
2x + 3y = 7 (equ 1)
3x + 2y = 8 (equ 2)

equation 1 times 3
6x + 9y = 21 (equ 1b)

Equation 2 times 2
6x + 4y = 16 (equ 2b)

Equation 1b - equation 2b

(6x + 9y) - (6x + 4y) = 21 - 16
6x + 9y - 6x - 4y = 5
5y = 5
Therefore y = 1 and from that you can work out x = 2.

That's easier than having: y = (8 - 3x)/2 etc

Maths teacher

there are several different ways to solve these.

if it helps they are called simultaneous equations.

one way is to draw both equations as graphs and see where they cross.

there’s other ways using algebra or bar modelling.

Just look up simultaneous equations. These are simple to solve.

He shouldn't need help from internet strangers.

He should know about simultaneous equations and apply his learned knowledge.