Question

four times a number and 3 times a number added together make 43 .two times the second number ,subtracted from three times the first give 11.what are the numbers?

Answer 1

Solutions=>>>>>>>>>>>

Let the first number be x
And second number be y

4x + 3y = 43 ______(i)
3x - 2y = 11 _______(ii)

[4x + 3y = 43] ×2
[3x - 2y = 11] × 3

= 8x + 6y = 86
9x - 6y = 33

17x = 119
=> x = 119/17 = 7

Putting the value of x in (i)

4x + 3y = 43
=> 4×7 + 3y = 43
=> 28 + 3y = 43
=> 3y = 43 - 28
=> y = 15/3 = 5

First number = 7
Second number = 5

Answer 2

$<b><i>Solutions ➠$

Let the first number be x
And second number be y

So,
The equation is =>
4x + 3y = 43 ______(i)
3x - 2y = 11 ______(ii)


Here,
let us eliminate the y term, and in order to eliminate the y term, equate the coefficient of y in both the equations.

[4x + 3y = 43] ×2
[3x - 2y = 11] × 3


Add equation (i) and (ii)

=> (8x + 6y) + (9x - 6y) = 86+33

=> 8x + 9x + 6y - 6y = 119

=> 17 x = 119

=> x = 119/17 = 7

Substitute the value of x in (i) we get,

4x + 3y = 43

=> 4×7 + 3y = 43

=> 28 + 3y = 43

=> 3y = 43 - 28

=> y = 15/3 = 5

Hence,
First number is 7
Second number is 5