Twice a number minus a second number is -1. Twice the second number added to three times the first number is 9. How do you find the two numbers?

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Answer 1 · 2016-12-06T06:21:07

The first number is $1$ $1$ and the second number is $3$ $3$ .

We consider the first number as $x$ $x$ and the second as $y$ $y$ . From the data, we can write two equations:

$2 x - y = - 1$ $2x-y=-1$
$3 x + 2 y = 9$ $3x+2y=9$

From the first equation, we derive a value for $y$ $y$ .

$2 x - y = - 1$ $2x-y=-1$

Add $y$ $y$ to both sides.

$2 x = - 1 + y$ $2x=-1+y$

Add $1$ $1$ to both sides.

$2 x + 1 = y$ $2x+1=y$ or $y = 2 x + 1$ $y=2x+1$

In the second equation, substitute $y$ $y$ with $(2 x + 1)$ $color(red)((2x+1))$ .

$3 x + 2 (2 x + 1) = 9$ $3x+2color(red)((2x+1))=9$

Open the brackets and simplify.

$3 x + 4 x + 2 = 9$ $3x+4x+2=9$

$7 x + 2 = 9$ $7x+2=9$

Subtract $2$ $2$ from both sides.

$7 x = 7$ $7x=7$

Divide both sides by $7$ $7$ .

$x = 1$ $x=1$

In the first equation, substitute $x$ $x$ with $1$ $color(red)1$ .

$(2 \times 1) - y = - 1$ $(2xxcolor(red)1)-y=-1$

$2 - y = - 1$ $2-y=-1$

Add $y$ $y$ to both sides.

$2 = y - 1$ $2=y-1$

Add $1$ $1$ to both sides.

$3 = y$ $3=y$ or $y = 3$ $y=3$