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Let
x represent one number and let
y represent the other number. Four times a first number decreased by a second number is -16 . The first number increased by twice the second number is -13 . Use the given conditions to write a system of equations. Solve the system and find the numbers.
Write a system of equations.
A.
{[4x-y=-16],[x+y=-14]:}
B.
{[4x-y=-16],[x+2y=-13]:}
C.
{[5x+y=-14],[x+y=-15]:}
D.
{[x-5y=-15],[x-y=-13]:}
Solve the system and find the numbers.
The numbers are
x=
◻ and
y=
◻

◻

Let $\mathrm{x}$ represent one number and let $\mathrm{y}$ represent the other number. Four times a first number decreased by a second number is $-16$ . The first number increased by twice the second number is $-13$ . Use the given conditions to write a system of equations. Solve the system and find the numbers. $\newline$ Write a system of equations. $\newline$ A. $\left\{\begin{array}{l}4 x-y=-16 \\ x+y=-14\end{array}\right.$ $\newline$ B. $\left\{\begin{array}{l}4 x-y=-16 \\ x+2 y=-13\end{array}\right.$ $\newline$ C. $\left\{\begin{array}{l}5 x+y=-14 \\ x+y=-15\end{array}\right.$ $\newline$ D. $\left\{\begin{array}{l}x-5 y=-15 \\ x-y=-13\end{array}\right.$ $\newline$ Solve the system and find the numbers. $\newline$ The numbers are $\mathrm{x}=$ $\square$ and $\mathrm{y}=$ $\square$ $\newline$ $\square$

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Solve one-step multiplication and division equations: word problems

Full solution

Q. Let

\mathrm{x}

represent one number and let

\mathrm{y}

represent the other number. Four times a first number decreased by a second number is

-16

. The first number increased by twice the second number is

-13

. Use the given conditions to write a system of equations. Solve the system and find the numbers.

\newline

Write a system of equations.

\newline

A.

\left\{\begin{array}{l}4 x-y=-16 \\ x+y=-14\end{array}\right.

\newline

B.

\left\{\begin{array}{l}4 x-y=-16 \\ x+2 y=-13\end{array}\right.

\newline

C.

\left\{\begin{array}{l}5 x+y=-14 \\ x+y=-15\end{array}\right.

\newline

D.

\left\{\begin{array}{l}x-5 y=-15 \\ x-y=-13\end{array}\right.

\newline

Solve the system and find the numbers.

\newline

The numbers are

\mathrm{x}=

\square

and

\mathrm{y}=

\square

\newline

\square

Multiply and Align $x$ Terms: Now, let's solve the system using substitution or elimination. I'll use elimination. $\newline$ First, I'll multiply the second equation by $4$ to align the $x$ terms: $\newline$ $4(x + 2y) = 4(-13)$ $\newline$ $4x + 8y = -52$
Subtract to Eliminate x: Next, I'll subtract the first equation from the modified second equation to eliminate x: $\newline$ $(4x + 8y) - (4x - y) = -52 - (-16)$ $\newline$ $4x + 8y - 4x + y = -52 + 16$ $\newline$ $8y + y = -36$ $\newline$ $9y = -36$
Solve for y: Now, I'll divide both sides by $9$ to solve for y: $\newline$ $\frac{9y}{9} = \frac{-36}{9}$ $\newline$ $y = -4$
Substitute for x: With the value of $y$ , I'll substitute it back into one of the original equations to solve for $x$ . I'll use the second equation: $\newline$ $x + 2(-4) = -13$ $\newline$ $x - 8 = -13$
Solve for x: Now, I'll add $8$ to both sides to solve for x: $\newline$ $x - 8 + 8 = -13 + 8$ $\newline$ $x = -5$
Check Solution: Let's check the solution by plugging the values of $x$ and $y$ into both original equations: $\newline$ First equation: $4(-5) - (-4) = -16$ $\newline$ Second equation: $-5 + 2(-4) = -13$
Check Solution: Let's check the solution by plugging the values of $x$ and $y$ into both original equations: $\newline$ First equation: $4(-5) - (-4) = -16$ $\newline$ Second equation: $-5 + 2(-4) = -13$ Checking the first equation: $\newline$ $4(-5) - (-4) = -20 + 4 = -16$ $\newline$ This matches the first equation. $\newline$ Checking the second equation: $\newline$ $-5 + 2(-4) = -5 - 8 = -13$ $\newline$ This matches the second equation.

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