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Introduction to systems with three variables

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Sal discusses how we approach solving systems of three variables algebraically, and how we visualize it graphically.

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Use the following problems as practice. Do the work, then check your answer.

1. −7x + 2y = 18 6x + 6y = 0 Answer: (−2, 2) 2. −16 + 20x − 8y = 0 36 = −18y − 22x Answer: (0, −2)

3. 0 = −2y + 10 − 6x 14 − 22y = 18x Answer: (2, −1)

4. −9 + 5y = −4x −11x = −20 + 9y Answer: (1, 1)

5. −8x − 10y = 24 6x + 5y = 2 Answer: (7, −8)

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When three variables are present, remember to solve/eliminate the same variable for each equation.

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The point of intersection of the two lines is the solution to the system.

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Always plug your solutions back into both of the original equations to make sure you have the correct answer.

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Solve for x and y.

2x - y = 9

3x + 4y = -14

The goal of elimination is to cancel out a variable, then combine the two equations into a single variable equation. To solve this system by elimination, you need to multiply the entire top equation by four so that -4y and +4y will cancel out.

8x - 4y = 36

3x + 4y = -14

The -4y and +4y cancel out.

8x = 36

3x = -14

Add down to combine the equations.

11x = 22

x = 2

Plug in your solution for x into the first equations, and isolate y.

2( 2 ) - y = 9

4 - y = 9

-y = 5

y = -5

State your solution as ( 2 , -5 ).

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Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations and solve. And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is its y-intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope-intercept form for us. So it's rise over run. So for every time it moves 5 to the right it's, going to move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the right, we will move up 3. So we will go right over there, and it looks like they intersect right at that point, right at the point x is equal to 5, y is equal to 2. So we'll type in x is equal to 5, y is equal to 2. And you could even verify by substituting those values into both equations, to show that it does satisfy both constraints. So let's check our answer. And it worked.

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Systems of equations with graphing: y=7/5x-5 & y=3/5x-1 (video) | Khan Academy

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Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations and solve.

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Solve for x and y. x - y = 2

4x - 6y = 4

One way to solve this system of equations is by deriving an expression for x from the first equation, and substituting it back into the second equation. Begin by adding y to both sides of the first equation. You now have:

x = y + 2

Substitute ( y + 2 ) in for x in the second equations. Isolate y.

4( y + 2 ) - 6y = 4

4y + 8 - 6y = 4

-2y = -4

y = 2

Now that you have the solution for y, plug in the solution into the first equation, and isolate y.

x - (2) = 2

x = 4

Your solutions can be written as

( 4 , 2 ).