# Thread: Need help with Elimination and substitution been stuck on this question

1. ## Need help with Elimination and substitution been stuck on this question

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. ## Re: Need help with Elimination and substitution been stuck on this question

Originally Posted by asdf1243
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. multiply each term in the first equation by 4 and the second equation by 3 and use elimination
3x-4y=10

y=3x-4. Substitute 3x-4 for y in the second equation. Solve for x, then y
2x+3y=10

x+4y=1. Substitute (1-4y) for x in the second equation, or multiply each term in the first
2x+5y=-1. equation by -2 and use elimination
2nd one: You are literally given a substitution for $y$. You may as well use it.
3rd one: It is easy to eliminate $x$ by taking the second equation minus twice the first one. It is equally easy to take the first equation, subtract $4y$ from both sides to isolate $x$ and substitute.