Question

solve each system of equations by elimination. clearly identify your solution.

1. \

$$\begin{cases} x + y = -5 \\\\ x - y = 9 \\end{cases}$$

1. \

$$\begin{cases} x + 5y = 20 \\\\ 2x - 7y = -45 \\end{cases}$$

1. \

$$\begin{cases} 4x + 3y = -1 \\\\ 5x + 4y = 1 \\end{cases}$$

1. \

$$\begin{cases} x = 2y - 3 \\\\ 2x - 3y = -5 \\end{cases}$$

© gina wilson (all things algebra®, llc), 2012 - 2016

Explanation:

Problem 9

Step1: Add the two equations

$$(x + y) + (x - y) = -5 + 9$$
$$2x = 4$$

Step2: Solve for $x$

$$x = \frac{4}{2} = 2$$

Step3: Substitute $x=2$ into first equation

$$2 + y = -5$$

Step4: Solve for $y$

$$y = -5 - 2 = -7$$

Problem 10

Step1: Multiply first equation by 2

$$2(x + 5y) = 2 \times 20$$
$$2x + 10y = 40$$

Step2: Subtract second equation

$$(2x + 10y) - (2x - 7y) = 40 - (-45)$$
$$17y = 85$$

Step3: Solve for $y$

$$y = \frac{85}{17} = 5$$

Step4: Substitute $y=5$ into first equation

$$x + 5(5) = 20$$
$$x + 25 = 20$$

Step5: Solve for $x$

$$x = 20 - 25 = -5$$

Problem 11

Step1: Multiply first equation by 4, second by 3

$$4(4x + 3y) = 4(-1) \implies 16x + 12y = -4$$
$$3(5x + 4y) = 3(1) \implies 15x + 12y = 3$$

Step2: Subtract the two new equations

$$(16x + 12y) - (15x + 12y) = -4 - 3$$
$$x = -7$$

Step3: Substitute $x=-7$ into first equation

$$4(-7) + 3y = -1$$
$$-28 + 3y = -1$$

Step4: Solve for $y$

$$3y = 27 \implies y = \frac{27}{3} = 9$$

Problem 12

Step1: Substitute $x=2y-3$ into second equation

$$2(2y - 3) - 3y = -5$$

Step2: Expand and simplify

$$4y - 6 - 3y = -5$$
$$y - 6 = -5$$

Step3: Solve for $y$

$$y = -5 + 6 = 1$$

Step4: Substitute $y=1$ into $x=2y-3$

$$x = 2(1) - 3 = 2 - 3 = -1$$

Answer:

1. $x=2$, $y=-7$
2. $x=-5$, $y=5$
3. $x=-7$, $y=9$
4. $x=-1$, $y=1$