Question

explain how you would use multiplication to solve each of these linear systems by elimination.

1. \\(\

$$\begin{cases}2x + 3y = 54\\\\4x + 7y = 85\\end{cases}$$

\\) \t\t3. \\(\

$$\begin{cases}3x + 2y = 54\\\\4x + 5y = 85\\end{cases}$$

\\) \t\t4. \\(\

$$\begin{cases}2x + 3y = 54\\\\-4x + 7y = 85\\end{cases}$$

\\)

Explanation:

Problem 2:

Step1: Multiply first equation by -2

To eliminate \(x\), we multiply the first equation \(2x + 3y = 54\) by \(-2\) to get \(-4x - 6y = -108\).

Step2: Add to second equation

Now add this new equation to the second equation \(4x + 7y = 85\):
\[

$$\begin{align*} (-4x - 6y) + (4x + 7y) &= -108 + 85\\ (-4x + 4x) + (-6y + 7y) &= -23\\ y &= -23 \end{align*}$$

\]

Step3: Substitute \(y\) back

Substitute \(y = -23\) into the first original equation \(2x + 3(-23) = 54\):
\[

$$\begin{align*} 2x - 69 &= 54\\ 2x &= 54 + 69\\ 2x &= 123\\ x &= \frac{123}{2} = 61.5 \end{align*}$$

\]

Problem 3:

Step1: Multiply first equation by -4 and second by 3

To eliminate \(x\), multiply the first equation \(3x + 2y = 54\) by \(-4\) to get \(-12x - 8y = -216\), and multiply the second equation \(4x + 5y = 85\) by \(3\) to get \(12x + 15y = 255\).

Step2: Add the two new equations

Add \(-12x - 8y = -216\) and \(12x + 15y = 255\):
\[

$$\begin{align*} (-12x - 8y) + (12x + 15y) &= -216 + 255\\ (-12x + 12x) + (-8y + 15y) &= 39\\ 7y &= 39\\ y &= \frac{39}{7} \end{align*}$$

\]

Step3: Substitute \(y\) back

Substitute \(y = \frac{39}{7}\) into the first original equation \(3x + 2(\frac{39}{7}) = 54\):
\[

$$\begin{align*} 3x + \frac{78}{7} &= 54\\ 3x &= 54 - \frac{78}{7}\\ 3x &= \frac{378 - 78}{7}\\ 3x &= \frac{300}{7}\\ x &= \frac{100}{7} \end{align*}$$

\]

Problem 4:

Step1: Add the two equations

Notice that the coefficients of \(x\) are \(2\) and \(-4\). If we multiply the first equation \(2x + 3y = 54\) by \(2\), we get \(4x + 6y = 108\). Now add this to the second equation \(-4x + 7y = 85\):
\[

$$\begin{align*} (4x + 6y) + (-4x + 7y) &= 108 + 85\\ (4x - 4x) + (6y + 7y) &= 193\\ 13y &= 193\\ y &= \frac{193}{13} \end{align*}$$

\]

Step2: Substitute \(y\) back

Substitute \(y = \frac{193}{13}\) into the first original equation \(2x + 3(\frac{193}{13}) = 54\):
\[

$$\begin{align*} 2x + \frac{579}{13} &= 54\\ 2x &= 54 - \frac{579}{13}\\ 2x &= \frac{702 - 579}{13}\\ 2x &= \frac{123}{13}\\ x &= \frac{123}{26} \end{align*}$$

\]

Answer:

s:

1. \(x = 61.5\), \(y = -23\)
2. \(x = \frac{100}{7}\), \(y = \frac{39}{7}\)
3. \(x = \frac{123}{26}\), \(y = \frac{193}{13}\)