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Use the elimination method to solve for p and q.
Equation 1 quad -2p-3q=-4
Equation 2 quad -10 p-21 q=-(55)/(2)

Use the elimination method to solve for $p$ and $q$ . $\newline$ Equation $1 \quad -2 p-3 q=-4$ $\newline$ Equation $2 \quad -10 p-21 q=-\frac{55}{2}$

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Solve one-step multiplication and division equations: word problems

Full solution

Q. Use the elimination method to solve for

p

and

q

.

\newline

Equation

1 \quad -2 p-3 q=-4

\newline

Equation

2 \quad -10 p-21 q=-\frac{55}{2}

Write Equations: First, let's write down the system of equations we need to solve: $\newline$ Equation $1$ : $-2p - 3q = -4$ $\newline$ Equation $2$ : $-10p - 21q = -\left(\frac{55}{2}\right)$
Multiply Equations: To use the elimination method, we need to multiply the equations by appropriate numbers so that one of the variables will be eliminated when we add or subtract the equations from each other. Let's find a common multiple for the coefficients of $p$ or $q$ in both equations.
Eliminate Variable: Looking at the coefficients of $p$ , which are $-2$ and $-10$ , we can multiply the first equation by $5$ to get the coefficient of $p$ in the first equation to match the coefficient of $p$ in the second equation in terms of absolute value. $\newline$ So, multiplying the first equation by $5$ , we get: $\newline$ $5(-2p - 3q) = 5(-4)$ $\newline$ $-10p - 15q = -20$ $\newline$ Now we have: $\newline$ Equation $1$ (multiplied by $5$ ): $-10p - 15q = -20$ $\newline$ Equation $2$ : $-2$ $1$
Solve for q: Next, we will subtract Equation $2$ from the modified Equation $1$ to eliminate $p$ : $\newline$ $(-10p - 15q) - (-10p - 21q) = -20 - (-(\frac{55}{2}))$ $\newline$ This simplifies to: $\newline$ $-10p + 10p - 15q + 21q = -20 + \frac{55}{2}$ $\newline$ The $p$ terms cancel out, and we are left with: $\newline$ $6q = -20 + \frac{55}{2}$
Substitute q: Now we need to solve for $q$ . First, let's convert $-20$ to a fraction with a denominator of $2$ to combine it with $55/2$ : $-20 = -40/2$ So the equation becomes: $6q = -40/2 + 55/2$
Isolate $-2p$ : Combining the fractions, we get: $\newline$ $6q = (-40 + 55) / 2$ $\newline$ $6q = 15 / 2$ $\newline$ Now, to solve for $q$ , we divide both sides by $6$ : $\newline$ $q = (15 / 2) / 6$ $\newline$ $q = 15 / 12$ $\newline$ $q = 5 / 4$
Solve for p: Now that we have the value of $q$ , we can substitute it back into one of the original equations to solve for $p$ . Let's use Equation $1$ : $\newline$ $-2p - 3(\frac{5}{4}) = -4$ $\newline$ $-2p - \frac{15}{4} = -4$ $\newline$ First, let's convert $-4$ to a fraction with a denominator of $4$ to combine it with $-\frac{15}{4}$ : $\newline$ $-4 = -\frac{16}{4}$ $\newline$ So the equation becomes: $\newline$ $-2p - \frac{15}{4} = -\frac{16}{4}$
Solve for p: Now that we have the value of $q$ , we can substitute it back into one of the original equations to solve for $p$ . Let's use Equation $1$ : $\newline$ $-2p - 3\left(\frac{5}{4}\right) = -4$ $\newline$ $-2p - \frac{15}{4} = -4$ $\newline$ First, let's convert $-4$ to a fraction with a denominator of $4$ to combine it with $-\frac{15}{4}$ : $\newline$ $-4 = -\frac{16}{4}$ $\newline$ So the equation becomes: $\newline$ $-2p - \frac{15}{4} = -\frac{16}{4}$ Now we will isolate $-2p$ : $\newline$ $-2p = -\frac{16}{4} + \frac{15}{4}$ $\newline$ $-2p = \left(-16 + 15\right) / 4$ $\newline$ $-2p = -\frac{1}{4}$ $\newline$ To solve for $p$ , we divide both sides by $-2$ : $\newline$ $p = \left(-\frac{1}{4}\right) / -2$ $\newline$ $p = \frac{1}{8}$

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