# There are 2 numbers whose sum is 53. 3 times the smaller number is equal to 19 more than the larger number. Which are the numbers?

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assume that the fsmaller number =S and the larger number is=L

then, S+L= 53....... (1)

3 times the smaller number = 19 more than the larger number

==> 3 S = 19 + L ........ (2)

Now, combining equation (1) and (2) we have:

S+L = 53

3S = 19 + L ==> L = 3S -19

substitute in eqution (1),

S + (3S- 19) = 53

4S - 19 = 53

4S = 72 divide by 4

S = 72/4 = 18

and L = 3(18)- 19 = 35

The larger number is 35 and the smaller number is 18. Here is how to find this.

Let X be the smaller number and Y be the larger.

x + y = 53

3x = y + 19

Let us solve for y in the first equation.

y = 53 -x

Then let us substitute that for y in the second equation.

3x = 53 - x + 19

3x = -x + 72

4x = 72

**x = 18**

If x = 18

18 + y = 53

y = 35

Put a=the smaller number and b=the larger number.

Their sum is 53.

a+b=53 (1)

Now we'll write the constraint: 3 times smaller is equal to 19 more than larger.

3a = b+19

3a-b = 19 (2)

If we'll add (1) to (2), we'll have:

a+b+3a-b = 53+19

4a = 72

**a = 18**

If we'll substitute the value for a in equation (1), we'll get:

a+b = 53

18+b = 53

b = 53-18

**b = 35**

Let x and y be the 2 numbers . So by data x+y =53.....(1)

If x is smaller number, the 3x = y+19....(2) From (2) we get:

3x-y = 19....(3)

x+y = 53...(1)

(1)+(2) gives: 4x = 19+53 = 72.. Or x= 72/4 = 18 is the smaller number and y= 53-x = 53 -18 =35.