# Thread: College algebra.

1. ## College algebra.

I need help with a few practice problems:

1.) 8x^-2 - 2x^-1 + 1 = 0

2.) Value of y?
-2x + 3z = 17
x - 2y - z = -3
-4x - 3y - z = 19

3.) Tasty Bakery sells three kinds of muffins: chocolate chip muffins at 15 cents each, oatmeal muffins at 20 cents each, and cranberry muffins at 25 cents each. Charles buys some of each kind and chooses twice as many cranberry muffins as chocolate chip muffins. If he spends $2.50 on 12 muffins, how many oatmeal muffins did he buy? 2. Originally Posted by xbchgrl04x 3.) Tasty Bakery sells three kinds of muffins: chocolate chip muffins at 15 cents each, oatmeal muffins at 20 cents each, and cranberry muffins at 25 cents each. Charles buys some of each kind and chooses twice as many cranberry muffins as chocolate chip muffins. If he spends$2.50 on 12 muffins, how many oatmeal muffins did he buy?
Let
x = # of chocolate chip muffins
y = # of oatmeal muffins
z = # of cranberry muffins

The 1st sentence & the last question gives you the first equation:
.15x + .20y + .25z = 2.50

The 2nd sentence gives you the second equation:
z = 2x or 2x - z = 0

The 2nd sentence and last question gives you the final equation:
x + y + z = 12

So your system of equations would be
.15x + .20y + .25z = 2.50
2x - z = 0
x + y + z = 12

Now solve. (Look at your previous threads on systems of linear equations for hints.) The value of y will give us the answer to the problem.

01

3. Originally Posted by xbchgrl04x
I need help with a few practice problems:

1.) 8x^-2 - 2x^-1 + 1 = 0
If you multiply both sides of the equation by $x^2$, this becomes $8- 2x+ x^2= 0$, a quadratic equation. Which, in fac, can be factored easily.

2.) Value of y?
-2x + 3z = 17
x - 2y - z = -3
-4x - 3y - z = 19
There are many different ways to solve systems of equations. One way is this: if you subtract the third equation from the second, the "z"s cancel:
(x- 2y- z)- (-4x- 3y- z)= (x+ 4x)- (-2- 3)y - (1- 1)z= 5x+ 5y= -3-19= -21.

Now multiply the second equation by 3, to get 3x- 6y- 3z= -9 and add it to the first equation: (-2x+ 3z)+ (3x- 6y- 3z)= (-2+3)x+ (0- 6)y+ (3- 3)zx= x- 6y= 17- 9= 8.

We have reduced to the two equations 5x+ 5y= -21 and x- 6y= 8. To eliminate x from those, multiply the second equation by 5 and subtract from the first: (5x+ 5y)- (5x- 30y)= 36y= -21- 40= -61. y= -61/36.

Another way to solve these would be to solve either the second or third equation for z, say the second equation for z= x- 2y+ 3, and replace the z in the other two equations by that:

-2x+ 3z= -2x+ 3(x- 2y+ 3)= (-2+3)x+ 3x- 6y+ 9= x- 6y+ 9= 17 so x- 6y= 8, an equation we got before.

-4x- 3y- z= -4x- 3y- (x- 2y+ 3)= -5x- 5y- 3= 19 so -5x- 5y= 21, which is the same as 5x+ 5y= -21 that we got before.

Now we can solve x- 6y= 8 for x= 6y+ 8 and put that into the second equation: 5x+ 5y= 5(6y+ 8)+ 5y= 35y+ 40= -21 or 35y= -61 so y= -61/35.

3.) Tasty Bakery sells three kinds of muffins: chocolate chip muffins at 15 cents each, oatmeal muffins at 20 cents each, and cranberry muffins at 25 cents each. Charles buys some of each kind and chooses twice as many cranberry muffins as chocolate chip muffins. If he spends \$2.50 on 12 muffins, how many oatmeal muffins did he buy?
yeongil did a good job on this one.