Brain Block on HL Maths HW

Spook_ie

Ok, I'm a bit stuck on a question for my maths HomeWork. It is a form of simultaneous equation.
The question is as follows:

The sum of four times one number and three times the second number is 61.
If twice the first number less the second number is 13, find the numbers.

The answers in the back of my book are 7 and 10 however I keep getting the incorrect answer.

Any help would be much appreciated.

Glowworm4

If you call the first number x and the second number y then from the first equation you get:

4x + 3y = 61

and for the second you get:

2x - y = 13

it is minus as the question says twice the first i.e 2x, less the second i.e y is equal to 13.

So you end up with

4x + 3y = 61
2x - y = 13

From there you just work it out to get your x and y values.

Indiego

Glowworm4 wrote: »

If you call the first number x and the second number y then from the first equation you get:

4x + 3y = 61

and for the second you get:

2x - y = 13

it is minus as the question says twice the first i.e 2x, less the second i.e y is equal to 13.

So you end up with

4x + 3y = 61
2x - y = 13

From there you just work it out to get your x and y values.

4x + 3y = 61
(2x - y = 13) x2

4x + 3y = 61
4x - 2y = 26 (subtract)

5y = 35
y = 7

2x - 7 = 13
2x = 20
x = 10

x = 10, y = 7

Spook_ie

Thanks a million for the quick replies! I understand completely now!