Need help w/ solving systems of linear equations by substitution

Timmy

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Here are three problems that I can't figure out.

The instructions say: Solve each system by substitution. Tell whether the system has one solution, infinitely-many solutions, or no solution.

1. 7x + 2y = -13
. .-3x - 8y = 23

2. x - 2y - 1 = 0
. .y - 5x + 14 = 0

3. y = -8x - 37
. .x + 3y = 4
 
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Solve each system by substitution. Tell whether the system has one solution, infinitely-many solutions, or no solution.
To "solve by substitution", one figures out how to multiply the equations in order to get one of the variables to drop out. They started (in your textbook and in your classroom lecture) with examples that were easy, such as:

. . . . .\(\displaystyle 3x\, -\, 5y\, =\, 1\)

. . . . .\(\displaystyle 2x\, +\, 5y\, =\, 9\)

Here, clearly the y-terms will cancel when you "add down":

. . . . .\(\displaystyle 3x\, -\, 5y\, =\, 1\)
. . . . .\(\displaystyle 2x\, +\, 5y\, =\, 9\)
. . . . .---------------
. . . . .\(\displaystyle 5x\, +\, 0y\, =\, 10\)

...so x = 2, and then you back-solve to find the value of y.

In the more complicated examples in class, you saw that you need to create this cancellation yourself, through multiplying the equations by useful values.

1. 7x + 2y = -13
. .-3x - 8y = 23
In your exercise (1) above, you could cancel the x-terms by multiplying the first equation by 3 (to get +21x) and the second equation by 7 (to get -21x). But these numbers are fairly large, and it's simpler to just multiply the first equation by 4 (to get +8y), which will cancel off with the y-term of the second equation (which is already -8y). There is no "rule" which requires that you use this multiplication but, just as with finding the least common multiple and the lowest common denominator, it's usually handy to use the simpler, smaller values.

2. x - 2y - 1 = 0
. .y - 5x + 14 = 0
You can start by rearranging to the standard format:

. . . . .\(\displaystyle +1x\, -\, 2y\, =\,\). . \(\displaystyle \, 1\)

. . . . .\(\displaystyle -5x\, +\, 1y\, =\, -14\)

You could multiply the first equation by, say, 20 (to get +20x) and the second equation by 4 (to get -20x), but wouldn't it be simpler to multiply just the second equation by 2 (to get +2y, which will cancel off with the -2y that's already in the first equation)?

3. y = -8x - 37
. .x + 3y = 4
Start by rearranging to get the standard format. What do you see? What are your thoughts? What will you try? Where will this lead?

If you get stuck, please reply showing all of your thoughts and steps so far. Thank you! ;)
 
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