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 \(4x + 7y = 85\):

$$
LATEXBLOCK0
$$

Step3: Substitute \(y\) back

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

$$
LATEXBLOCK1
$$

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 \(12x + 15y = 255\):

$$
LATEXBLOCK0
$$

Step3: Substitute \(y\) back

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

$$
LATEXBLOCK1
$$

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 \(-4x + 7y = 85\):

$$
LATEXBLOCK0
$$

Step2: Substitute \(y\) back

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

$$
LATEXBLOCK1
$$

Answer:

s:

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