Question

A first number plus twice a second number is 14. Twice the first number plus the second totals 10. Find the numbers.

Answer 1

Answer:

first number(x) = 2 second number(y)= 6

Step-by-step explanation:

This is an example of a simultaneous equation.

First write this word problem as equations, where x is the "first number" that you've mentioned and y is the "second number".

x + 2y = 14  (equation 1)

2x + y = 10  (equation 2)

This is solved using the elimination method.

We need to make one of the coefficients the same - in this case we can make y the same. In order to do this we need to multiply equation 2 by 2, so that y becomes 2y.

2x + y = 10 MULTIPLY BY 2

4x + 2y = 20 (this is now our new equation 2 with the same y coefficient)

Now subtract equation 1 from equation 2.

4x - x + 2y - 2y = 20 - 14  (2y cancels out here)

3x = 6

x = 2

Now we substitute our x value into equation 1 to find the value of y.

2 + 2y = 14

2y = 12

y = 6

Hope this has answered your question.

Answer 2

Answer:

6 and 2

Step-by-step explanation:

Let the first number =a

Let the second number =b

A first number plus twice a second number is 14.

• a+2b=14

Twice the first number plus the second totals 10.

• 2a+b=10

We solve the two equations simultaneously

$a+2b=14 \implies a=14-2b\\ Substitute into the second equation \\2(14-2b)+b=10\\28-4b+b=10\\-3b=10-28\\-3b=-18\\b=6$

Recall:

a=14-2b

=14-2(6)

=14-12

a=2

The two numbers are 6 and 2.