Solve each system by elimination. In systems with fractions, first clear denominators. See Example 2.
4x + y = -23
x - 2y = -17

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Accepted Answer

Hey, everyone. Here we are asked to use the elimination method to solve the given system of equations where the first equation is six, X plus 15, Y is equal to negative 33. And the second equation is X minus Y is equal to negative nine. Here we have four answer choice options. Answer choice A X is equal to negative eight and Y is equal to one answer. B X is equal to zero and Y is equal to answer C X is equal to negative nine and Y is equal to six answer D no solution. So here utilizing the elimination method, we first need to choose one of the variables that we want to solve for. And so we can begin by solving for X. And this means that we first need to eliminate Y. So to eliminate Y, we need to get both Y terms coefficient to be the same regardless of their sign. So to get the Y terms to be the same, we see that the Y in our first equation has a coefficient of 15. And we can easily make the Y in the second equation have the same coefficient by simply multiplying the left side and the right side of this second equation by 15. So multiplying by 15, we get the equation 15, X minus 15, Y is equal to negative 135. And so now, just recalling our original first equation, we had six X plus 15 Y is equal to negative 33. And so now we see here that both Y terms have the same coefficient but they have signs. So to eliminate our Y terms, we just have to add the two equations together. And so we have six X plus 15 X which gives us 21 X, we then have 15 Y plus negative 15 Y which gives us zero. Like we intended to obtain, since we wanted to eliminate Y and then this is equal to the right hand side which is negative 33 plus negative 135. And this gives us negative 168. So now we obtained a third equation after adding our two equations together. And so we have 21 X is equal to negative 168. Now just solving for X, we just divide both sides by 21. And so we see that our value for X is negative eight. So now that we have solved for X, we can move on to solving four Y. And so recalling our original two first equations from our original system, we have six X plus 15, Y is equal to negative 33. And from our second equation, we have X minus Y is equal to negative nine. Well, here we need to eliminate our X terms since we want to solve for Y. And so that means we want our X terms to be the same coefficient. And so we can simply take our second equation and now multiply it by negative six since our coefficient for the first, first equation is positive six. So now multiplying by negative six, we get negative six, X plus six, Y is equal to 54. And recalling again, our first equation, we have six, X plus 15 Y is equal to negative 33. So now we see our X terms have the same coefficient but opposite signs. So again, to eliminate these terms, we need to add the equations together. So adding the equations, our X terms cancel and we get zero, we then have 15 Y plus six Y which gives us plus 21. Y, we then have negative 33 plus 54 which gives us 21. So we get our 30 equation which is 21. Y is equal to 21. Now solving for Y, we divide both sides by 21 we see Y is equal to one. So now we have found our values for X and Y and we are left with answer choice A where X is equal to negative eight and Y is equal to one. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each system by elimination. In systems with fractions, first clear denominators. See Example 2. 4x + y = -23 x - 2y = -17

Verified Solution

Key Concepts

Systems of Equations

Elimination Method

Clearing Denominators