The sum of two numbers is 4. Twice the larger is 11 more than the smaller. How do you find the smaller number?

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Answer 1 · 2017-01-05T08:58:19

The smaller number is $- 1$ $-1$ .

Considering the two numbers as $x$ $x$ and $y$ $y$ where $x$ $x$ is the larger number, we can write:

$x + y = 4$ $x+y=4$
$2 x = y + 11$ $2x=y+11$

From the first equation, we can determine a value for $x$ $x$ .

$x + y = 4$ $x+y=4$

Subtract $y$ $y$ from both sides.

$x = 4 - y$ $x=4-y$

In the second equation, substitute $x$ $x$ with $(4 - y)$ $color(red)((4-y))$ .

$2 x = y + 11$ $2x=y+11$

$2 (4 - y) = y + 11$ $2color(red)((4-y))=y+11$

Open the brackets and simplify. The product of a positive and a negative is a negative.

$8 - 2 y = y + 11$ $8-2y=y+11$

Add $2 y$ $2y$ to both sides.

$8 = 3 y + 11$ $8=3y+11$

Subtract $11$ $11$ from both sides.

$- 3 = 3 y$ $-3=3y$

Divide both sides by $3$ $3$ .

$- 1 = y$ $-1=y$ or $y = - 1$ $y=-1$

In the first equation, substitute $y$ $y$ with $- 1$ $color(blue)(-1)$ .

$x + y = 4$ $x+y=4$

$x + (- 1) = 4$ $x+(color(blue)(-1))=4$

Open the brackets and simplify. The product of a positive and a negative is a negative.

$x - 1 = 4$ $x-1=4$

Add $1$ $1$ to both sides.

$x = 5$ $x=5$