### About Solving Linear Systems in Two Variables:

We often need to solve a linear system in two variables. Two common algebraic methods are substitution and elimination. With substitution, we can solve one of the equations for one of the variables. We can then plug in for this variable in one of the original equations. This will give us a linear equation in one variable. Once we solve this equation, we can plug back into one of the original equations and find our other unknown. With the elimination method, we aim to change the equations such that one pair of variable terms are opposites. We want to make sure that each equation is in standard form (ax + by = c) and then we can add the left sides and set this equal to the sum of the right sides. We will find that one variable will drop out and we are left with a linear equation in one variable. From this, we can then solve and substitute back into one of the original equations to find our other unknown.

Test Objectives

- Demonstrate the ability to solve a linear system using substitution
- Demonstrate the ability to solve a linear system using elimination

#1:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$-8x - 7y=-28$$ $$ 3x + 8y=-11$$

$$b)\hspace{.2em}$$ $$8x - 3y=-12$$ $$ -6x + 2y=10$$

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#2:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$8x + 5y=6$$ $$ -6x - 4y=-4$$

$$b)\hspace{.2em}$$ $$-30x + 30y=0$$ $$ 40x - 40y=0$$

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#3:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$-6x - 2y=-22$$ $$ 4x + 7y=-8$$

$$b)\hspace{.2em}$$ $$-10x - 5y=-20$$ $$ 7x + 7y=14$$

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#4:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$64x + 56y=-4$$ $$ -24x - 21y=0$$

$$b)\hspace{.2em}$$ $$-9x - 5y=-29$$ $$ 2x + 6y=26$$

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#5:

Instructions: solve each system.

$$a)\hspace{.2em}$$ $$-7x + 6y=29$$ $$ -3x - 7y=22$$

$$b)\hspace{.2em}$$ $$5x + 10y=5$$ $$ -7x - 6y=17$$

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Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}(7,-4)$$

$$b)\hspace{.2em}(-3,-4)$$

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#2:

Solutions:

$$a)\hspace{.2em}(2,-2)$$

b) Infinite Number of Solutions

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#3:

Solutions:

$$a)\hspace{.2em}(5,-4)$$

$$b)\hspace{.2em}(2,0)$$

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#4:

Solutions:

a) No Solution

$$b)\hspace{.2em}(1,4)$$

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#5:

Solutions:

$$a)\hspace{.2em}(-5,-1)$$

$$b)\hspace{.2em}(-5,3)$$