Solve each problem. See Examples 5 and 9. Solve the system of ...

Question

Solve each problem. See Examples 5 and 9. Solve the system of equations (4), (5), and (6) from Example 9.

$25 x + 40 y + 20 z = 2200$ (4)

$4 x + 2 y + 3 z = 280$ (5)

$3 x + 2 y + z = 180$ (6)

Accepted Answer

Solve the given system of linear equations. We have three equations here which I'll label as equations 12 and three equation one plus five, X plus 10, Y plus 15, Z equals to 2200. Version two is X plus four, Y plus two, Z equals 8 80 equation three plus two, X plus Y plus two, Z equals 1100. No, we have four possible answers which are all just sets of numbers X Y and Z let's see if we can solve for these equations. We can do this by using elimination. So let's look at the equations one and two. First, we want to get rid of one of the variables. Let's say you want to get rid of X to do this. We're going to add these equations together and remove one of the variables by multiplying by a factor. For instance, let's say you multiply equation two by negative five and add to equation one. This should get rid of the exits equation one is five X plus 10, Y plus 15. Z equals 2200. If you multiply equation two by negative five, we get negative five X minus 20. What minus 10 equals negative 4400. Adding together cancels other XS giving us negative 10, Y plus five Z equals negative 2200. We can do this again with equations two and three. Let's get rid of the X again by multiplying equation two by negative two to multiply two by negative two. Add to equation three, we would get negative two, X minus eight, Y minus four, Z equals negative of 17 and two X plus Y plus two Z equals once it adding those together gives us negative seven, Y minus two, Z equals negative 16 50. Now we can set up a new system with new equations. 4th and 5th, we have negative 10, Y plus five, Z equals negative 2200 and negative seven Y minus two Z equals negative 16 50. Now we need to get rid of another variable. Let's say we get rid of our Z. Next, let's multiply equation four by two and supply equation five by five. Equation four. We would get negative 20. Why plus 10 equals negative 4400. Equation five. We, we get negative 35. Why minus equals negative 16 50 multiplied by five, which that turns out to be native 82 50. We can now add these together and we will end up getting rid of our seats. Thank you. This negative 55 Y equals negative 80 to 50. Now that's gives us negative 12 6 50. Like so now let's divide both sides by negative by both sides. On 18 55 we get Y equals negative or positive rather 230. Now we have Y equals 2 30. We can start substituting our value in two other equations. Let's say we substitute into equation five, say substitute why? Into equation five, we will get negative seven, multiplied by 2 30 minus two Z equals negative 16 50. So now we multiply 2 30 by negative seven to get negative 16, 10 minus two to C was negative 50 add 16 10 on both sides to get negative two, Z equals negative 40. Finally divided by negative two to get C equals 20. Finally, we have Y and Z which we can plug into a final equation. Let's study, plug it into equation two. We all say substitute Y and Z and two. Equation two. Our equation two is X plus four, Y plus two Z and that s equals to 8 M X plus four multiplied by 2 30 plus two, multiplied by 20 equals 8 80. X plus 9 20 plus equals 8 80 or X plus 9 60 equals 80. Finally, we subtract both sides by 9 60 to get X is equals to negative 80. We now have three answers. X equals negative 80. Y is 2 30 Z is 20. Our final answer is the set N 80 30 20. If we look at our possible answers, we see the answer to our problem. His answer c ok. I hope to help you solve the problem. Thank you for watching. Goodbye.

Solve each problem. See Examples 5 and 9. Solve the system of equations (4), (5), and (6) from Example 9.
$25 x + 40 y + 20 z = 2200$ (4)
$4 x + 2 y + 3 z = 280$ (5)
$3 x + 2 y + z = 180$ (6)

Key Concepts

Systems of Linear Equations

Methods for Solving Systems (Substitution, Elimination, and Matrix Methods)

Checking Solutions for Consistency

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