# Linear Algebra and Solving Systems of Linear Equations.

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1. Given the equation y+5 = -2(x-3). find the slope and a point on the line

2. Determine the equation of the line, in slope- intercept form that is parallel to 5x+4y = -3 and contains (-8,9)

3. Find the equation of the line through the point (-2,6) and perpendicular to y= (3/5)x -2

4. Use the graphing method to find the solution of the y=2x+5 following system, if it exists. y= 8-x

5. Use the substitution to solve the system and then enter the x-coordinate of the solution

x+2y=10

x=3y

6. Solve the system of linear equation by the substitution method

2x-y=8

y=6-5x

7. Solve the system by the elimination method

2x-6y=3

4x+9y= -1

8. Solve the system by the elimination method.

6x-9y=7

-4x+6y=10

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##### Solution Summary

Shows how to put an equation into Slope-Intercept Form. How to determine if two equations are parallel or perpendicular. As well as a step by step explanation of how to solve a system of two linear equation. It shows how to solve the system using the Elimination method, also known as the Addition method. Also shows how to solve using the Substitution method.

##### Solution Preview

1. Given the equation y+5 = -2(x-3). Find the slope and a point on the line.

Write the equation in slope intercept form y = mx + b

The slope is 'm' and a point on the line is the y-intercept (0,b).

Start by using the distributive property on the right side of equation. y+5=-2x+6

Then subtract 5 from both sides: y=-2x+1

Slope is -2, and the y-intercept is (0, 1)

2. Determine the equation of the line, in slope- intercept form that is parallel to 5x+4y = -3 and contains

(-8,9)

First write given equation in slope intercept form. y=(-5/4)x -3/4

If the second line is parallel, then they have the same slope, -5/4

Now use point slope form to get new equation: y-y1=m(x-x1)

Point (-8, 9) is the point (x1 , y1)

y-9=-5/4(x-(-8))

Distribute y-9=(-5/4)x-10

Add 9 to both sides: y=(-5/4)x -1

3. Find the equation of the line through the point (-2,6) and perpendicular to y= (3/5)x -2

For the lines to be perpendicular the ...

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