Question

half the sum of two numbers is 20 and three times their difference is 18. find the numbers (simultaneous linear equation)

Answer 1

Let the numbers be x and y.

Given that half the sum of two numbers = 20.

(x + y)/2 = 20

x + y = 20 * 2

x + y = 40.

y = 40 - x  ---- (1)


Given that three times their difference = 18.

3(y - x) = 18

y - x = 18/3

y - x = 6

40 - x  - x= 6

40 - 2x = 6

-2x = 6 - 40

-2x = - 34

x = 17


SUbstitute x = 17 in (1), we get

y = 40 - x

y = 40 - 17

y = 23.


Therefore the numbers are 17 and 23.



Verification:

(17 + 23)/2 = 20

40/2 = 20

20 = 20.


3(y - x) = 18

3(23 - 17) = 18

3(6) = 18

18 = 18.



Hope this helps!

Answer 2

Let the numbers be p and q

Given that half the sum of two numbers = 20.

(p+ q)/2 = 20

p+ q = 20 * 2

p + q= 40.

q = 40 - p



three times their difference = 18.

3(q- p) = 18

q- p = 18/3

q- p. = 6

40 - p - p =. 6

40 - 2p= 6

-2p= 6 - 40

-2p = - 34

p= 17


we get

q = 40 - p

q = 40 - 17

q = 23.


Therefore the numbers are 17 and 23.



Verification:

(17 + 23)/2 = 20

40/2 = 20

20 = 20.


3(q- p) = 18

3(23 - 17) = 18

3(6) = 18

18 = 18.

p q = 18

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