Answers

2014-11-28T12:37:52+08:00
Let the numbers be x and y

x be the larger; y be the smaller

x + y = 23

Double the larger the number exceeds three times the smaller by sixteen

2x = 3y + 16

Let's rearrange the first equation to find the value of x

x + y = 23

Transpose y

x = 23 - y

Now substitute the value of x to the 2nd equation

2x = 3y + 16

2(23 - y) = 3y + 16

Multiply

46 - 2y = 3y + 16

Transpose 3y and 46

-2y - 3y = 16 - 46

Add like sign and subtract unlike sign

-5y = -30

Divide by -5 both sides

-5y/-5 = -30/-5

Therefore

y = 6

Let us now find for x

x = 23 - y

Substitute value of y

x = 23 - 6

Therefore

x = 17

Let us check by substituting the value of x and y to the second equation

2x = 3y + 16

2(17) = 3(6) + 16

Multiply

34 = 18 + 16

Add

34 = 34

I hope it helps
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