# Thread: urgent algebra help needed!

1. ## urgent algebra help needed!

Linear Systems question - solve this system:

1/x + 3/y = 3/4
3/x - 2/y = 5/12

Polynomials question:

Divide 2x – 1 into 4x4 – 5x2 + 2x + 4.

2. Originally Posted by 14041471
Linear Systems question - solve this system:

1/x + 3/y = 3/4
3/x - 2/y = 5/12

Polynomials question:

Divide 2x – 1 into 4x4 – 5x2 + 2x + 4.
For the first question, multiply the first equation by -3 and then add it to the second equation:

-3(1/x + 3/y = 3/4)
+ 3/x - 2/y = 5/12

-3/x - 9/y = -9/4
+3/x - 2/y = 5/12 Find a common denominator for the right hand sides of the equations:

-3/x - 9/y = -27/12
+3/x - 2/y = 5/12 Multiplied the right hand side of the top equation by 3/3, which is essentially multiplying by 1.

-11/y = -22/12

Cancel out the negatives and cross multiply to get:
12*11 = 22*y
132 = 22y
y = 6

Substitute this value back into this original equation:
3/x - 2/y = 5/12 to find x.

3/x - 2/6 = 5/12
3/x = 5/12 + 4/12 got a common denominator for the constants
3/x = 9/12

Cross Multiply:
3*12 = 9*x
36 = 9x
x = 4

For practice, Substitute the x and y values back into both of the original equations to show that they check out.

3. Originally Posted by 14041471
Linear Systems question - solve this system:

1/x + 3/y = 3/4
3/x - 2/y = 5/12

Polynomials question:

Divide 2x – 1 into 4x4 – 5x2 + 2x + 4.
Multiply the first equation by 2, the second equation by 3:

1/x + 3/y = 3/4 ==> 2/x + 6/y = 6/4

3/x - 2/y = 5/12 ==> 9/x-6/y = 15/12

Add columnwise and you'll get 11/x = 11/4 ==> x = 4

Re-substitute and calculate y: y = 6

4. ## 2nd question

Originally Posted by 14041471
Linear Systems question - solve this system:

1/x + 3/y = 3/4
3/x - 2/y = 5/12

Polynomials question:

Divide 2x – 1 into 4x4 – 5x2 + 2x + 4.
For the polynomials question, you start by asking yourself how many times 2x can go into 4x^4. Well, it would be 2x^3, because 2x*2x^3 = 4x^4.

So you multiply (2x-1) by 2x^3 and subtract it from the entire above polynomial: Looks like:

2x^3(2x-1) = 4x^4 - 2x^3

Now, do: (4x^4 - 5x^2 + 2x + 4) - (4x^4 - 2x^3)
The polynomial now looks like:
ii) 2x^3 - 5x^2 + 2x + 4

We do the same thing as above, ask ourselves how many times 2x can go into 2x^3. it would be x^2 times, since 2x*x^2 = 2x^3

So you multiply (2x-1) by x^2 and subtract it from the second stage polynomial (ii) above:

(x^2)(2x-1) = 2x^3 - x^2

Now, do (2x^3 - 5x^2 + 2x + 4) - (2x^3 - x^2)
The polynomial now looks like:
(iii) -4x^2 + 2x + 4

We do the same, and se that 2x can go into -4x^2 a number of -2x times because 2x*-2x = -4x^2.

So, you multiply (2x-1) by -2x and subtract it from the third stage polynomial (iii)

-2x(2x - 1) = -4x^2 + 2x

Now, do (-4x^2 + 2x + 4) - (-4x^2 + 2x)
The polynomial now looks like:
(iv) 4

You can't divide 2x into 4 any further, so we are done. The ending form looks like:

(2x-1)(2x^3 + x^2 -2x) + 4

Expand this to check to see if you get the original polynomial at the beginning. Hope this helps