Solve the equation (2x+3)/(x-4)-(2x-8)/(2x+1)=1 and give the answer to 2 decimal places

To start off with we need to remove the two denominators (the bottom part of the fractions) on the left hand side of the equation. This can be achieved by multiplying the through the whole equation by (x-4)(2x+1), since (x-4)/(x-4)=1 and (2x+1)/(2x+1)=1. This procedure transforms the equation into the following:

(2x+3)(2x+1)-(2x-8)(x-4)=(x-4)(2x+1)

Now we continue by multiplying out the brackets, which gives:

(4x^2+8x+3)-(2x^2-16x+32)=(2x^2-7x-4)

Which simplifies to:

24x-29=-7x-4 (remember when subtracting the left hand side, that minus a minus number is a plus. Also note that we want to add and subtract according to the power of the x variable i.e. the x^2 needs to be in a separate grouping to the x^1 and x^0).

Continue with the equation by adding 7x to each side of the equality and adding 29 to each side. This yields the following:

31x=25

Now to get the solution for the variable x, we divide both sides by 31 and this gives:

x=25/31

Inputting this fraction into the calculator gives the decimal solution as:

0.8064516129

The question asks for the answer to be to decimal places and so the answer is:

0.81

Overview: This is hard question for a GCSEs maths student and is in the A* ability criteria, but the through some simple techniques/ steps, the question is a lot more manageable and solvable.

A breakdown of the steps:

1. Multiply the whole of the equation by the product of the denominators of the fractions. There by eliminating the denominators.

2. Expand the brackets with care.

3. Group the correct coefficients when adding/subtracting {i.e. keeping the powers of x together).

4. Once the coefficients have been grouped together successfully, then manipulate the equation by subtracting/adding on each side to get x equal to a number.

5. Remember to put the answer in the form wanted by the question. If it does not state the form wanted then put it into 2 decimal places to be safe.

Answered by Jake J. Maths tutor

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