Use the substitution or elimination method to solve each syste...

Question

Use the substitution or elimination method to solve each system of equations. Identify any inconsistent systems or systems with infinitely many solutions. If a system has infinitely many solutions, write the solution set with y arbitrary.
6x + 10y = -11
9x + 6y = -3

6x + 10y = -11

9x + 6y = -3

Accepted Answer

Hello everybody. I have everything right today. Today we're going to be looking at this math question that states solve the given system of linear equations using the elimination method. And the system of equations includes 12 X plus eight, Y equals negative 15 and five X plus three Y equals negative two. Now the s provided are a 2029 divided by four comma negative 51 divided by four B 29 divided by four comma 51 divided by four C negative 29 divided by four comma negative 51 divided by four or D no solution. Now, when we have a system of linear equations like this and especially using the elimination method, we have to think about what can we, how can we manipulate each of the equations to try to cancel out one of the variables? So in this case, I'm going to label equation. So I'm going to put 12 X plus eight Y equals negative 15. I'm going to label that equation one and five X plus three Y equals negative two. That'll be equation two. So to eliminate a variable here, I'm going to multiply and in this case it's specifically to eliminate variable Y. I'm going to multiply equation one by negative three, an equation two by positive eight. So I'll do exactly that. So I have negative three and I'm going to place each equation one on top of the other. And I'll explain a little bit further on that a little later on, but that's how I'm going to write it and you'll see why. So I'll have negative three, multiplying 12 X plus eight, Y equals negative and positively multiplying five X plus three, Y equals negative two. And we have to keep in mind that we are multiplying the entire thing including the answers. So we'll have negative three, multiplying 12 X is negative 36 X and then negative three, multiplying a positive eight Y. So we'll have minus 24 Y and that equals a negative 15, multiply negative three which will be a positive of 45. And I'm going to place that equation on top and under it, I'll be placing equation two which is going to be eight multiplying five X which is 40 X and then a multiplying a positive three Y which will be plus 24 Y and that will be equal to a negative two, multiplying that positive A which is negative 16. Now I placed them one on top of the other. So once you write it like that, we think about adding each equation up. So this is how we would write or when we're doing Normal edition, when we learn addition, we learn to write the numbers one on top of the other place our, our positive sign and draw a line under it and just add. So this is what, how we have to think about this. We're just adding the equations. So negative 36 X plus 40 X will be positive for X negative 24 plus positive 24 will be zero. So we'll have that four, X is equal to 45 minus 16, which is 29. And now all we have to do is solve for X. So we have that if we divide both sides by four, we have that X is equal to 29 divided by four. So that's our first answer for X. And with that, we can plug that back in into equation one or equation two. But I'll be using equation one, two salt for the Y. So we had 12 X plus eight, Y equals negative 15. So once we plug in that X, we'll have 12 multiplying one in nine divided by four plus A Y equals negative 15. Now we have a number multiplying a fraction. So we multiply the 12 with the numerator of that fraction. So 12 multiplying 29 which is 348 and that'll still be over four or divided by four. And that'll be plus eight, Y equals negative 15 and 348 divided by four is 87. So I have 87 plus eight, Y equals negative 15. And we can move that 87 over from the left hand side to the right hand side and we'll have eight, Y equals negative 102. And now we can't eight does not divide into 100 and two into and two perfectly. So we'll have a fraction of one oh two divided by eight, which can be simplified a bit. So we'll have Y is equal to negative 51 divided by four. So all we did to simplify, there was just divide both the numerator and the denominator by two. And that will be our answer for Y. So we have X equals 29 divided by four and Y equals negative 51 divided by four or in this case, answer choice. A 29 divided by four comma negative 51 divided by four. So I hope that was helpful. And until next time.

Use the substitution or elimination method to solve each system of equations. Identify any inconsistent systems or systems with infinitely many solutions. If a system has infinitely many solutions, write the solution set with y arbitrary.
6x + 10y = -11
9x + 6y = -3

Key Concepts

Substitution and Elimination Methods

Types of Solutions for Systems of Equations

Expressing Solutions with Parameters

Watch next

Use the substitution or elimination method to solve each system of equations. Identify any inconsistent systems or systems with infinitely many solutions. If a system has infinitely many solutions, write the solution set with y arbitrary. 6x + 10y = -11 9x + 6y = -3

Key Concepts

Substitution and Elimination Methods

Types of Solutions for Systems of Equations

Expressing Solutions with Parameters

Watch next

Use the substitution or elimination method to solve each system of equations. Identify any inconsistent systems or systems with infinitely many solutions. If a system has infinitely many solutions, write the solution set with y arbitrary.
6x + 10y = -11
9x + 6y = -3