# Thread: Applications of Linear Systems

1. ## Applications of Linear Systems

I'm stuck with following question :S

Systems of equations can be solved algebraically by using the substitution method and the elimination method. Sometimes, one method is a better than the other. At other times, it is just as easy to use either method. For each system displayed below, decide whether it would be easier to solve the system using the substitution method, the elimination method or either method. Provide an explanation to support why you made your decision.

2x + 3y = 1
3x - 4y = 10

y = 3x - 4
2x + 3y = 10

x + 4y = 1
2x + 5y = -1

2. Originally Posted by Serialkisser
I'm stuck with following question :S

Systems of equations can be solved algebraically by using the substitution method and the elimination method. Sometimes, one method is a better than the other. At other times, it is just as easy to use either method. For each system displayed below, decide whether it would be easier to solve the system using the substitution method, the elimination method or either method. Provide an explanation to support why you made your decision.

2x + 3y = 1
3x - 4y = 10

y = 3x - 4
2x + 3y = 10

x + 4y = 1
2x + 5y = -1
For the first one, I'd probably choose elimination....

$\displaystyle 2x+3y=1$
$\displaystyle 3x-4y=10$

Multiply the first equation by 4 and the second equation by 3 to get:

$\displaystyle 8x+12y=4$
$\displaystyle 9x-12y=30$

Now, add the two equations to get:

$\displaystyle 17x=34$
$\displaystyle \boxed{x=2}$

Substitute x=2 back into either equation to solve for y.

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The second system looks like a good candidate for the substitution method since the first equation is already solved for y.

$\displaystyle y = 3x - 4$
$\displaystyle 2x + 3y = 10$

Substitute 3x-4 for y in the second equation to get:

$\displaystyle 2x+3(3x-4)=10$

Solve for x. Then use the first equation to find y.
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Your last system could go either way, elimination or substitution. One is about as easy as the other.