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You could multiply the first equation by 3 and the second by -5 and eliminate the x... OR you could multiply the first equation by 7 and the second by 10 and eliminate the y. Either way works.

Q: How would you solve this using elimination 5x-10y equals 15 3x plus 7y equals 31?

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(2,-2)

By elimination: x = 3 and y = 0

You cannot solve one linear equation in two variables. You need two equations that are independent.

Yes and it works out that x = 3 and y = 4

4

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(2,-2)

By elimination: x = 3 and y = 0

You cannot solve one linear equation in two variables. You need two equations that are independent.

Yes and it works out that x = 3 and y = 4

4

Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.

You can solve lineaar quadratic systems by either the elimination or the substitution methods. You can also solve them using the comparison method. Which method works best depends on which method the person solving them is comfortable with.

No. Solving equations in two unknowns requires two independent equations. Since you have only one equation there is no solution.

Multiply every term in both equations by any number that is not 0 or 1, and has not been posted in our discussion already. Then solve the new system you have created using elimination or substitution method:6x + 9y = -310x - 6y = 58

Elimination is particularly easy when one of the coefficients is one, or the equation can be divided by a number to reduce a coefficient to one. This makes substitution and elimination more trivial.

X=0 and y=0

5x - 4y ≥ -203x - 2y ≤ -8y ≥ -3