# Video: CBSE Class X • Pack 3 • 2016 • Question 16

CBSE Class X • Pack 3 • 2016 • Question 16

06:35

### Video Transcript

Solve the equation one divided by 𝑥 minus one multiplied by 𝑥 minus two plus one divided by 𝑥 minus two multiplied by 𝑥 minus three is equal to two-thirds for 𝑥, given that 𝑥 is not equal to one, two, or three.

The reason that 𝑥 cannot be equal to one, two, or three is that when we substitute one of those values into the left-hand side of the equation, we end up with a denominator equal to zero. Dividing by zero is not possible as the answer is undefined. Therefore, 𝑥 cannot equal one, two, or three in this question. Knowing this won’t affect on method. But it will enable us to rule out some solutions at the end.

Our first step when dealing with algebraic fractions is to find a common denominator in the same way as we would with two ordinary fractions. Remember with fractions, whatever you do to the numerator, you must do to the denominator. In this case, we’ll multiply the top and the bottom of the first fraction by 𝑥 minus three. We’ll multiply the top and the bottom of the second fraction by 𝑥 minus one. The denominator on both fractions on the left-hand side of the equation is now the same: 𝑥 minus one multiplied by 𝑥 minus two multiplied by 𝑥 minus three.

When adding or subtracting two fractions with the same denominator, we just need to add the two numerators. In this case, we need to add 𝑥 minus three and 𝑥 minus one. This gives us two 𝑥 minus four divided by 𝑥 minus one multiplied by 𝑥 minus two multiplied by 𝑥 minus three is equal to two-thirds as 𝑥 plus 𝑥 is equal to two 𝑥 and negative three minus one is equal to negative four.

Our next step is to expand or multiply out the brackets on the denominator of the left-hand side. Firstly, we’ll multiply 𝑥 minus one by 𝑥 minus two using the FOIL method. Multiplying the first terms gives us 𝑥 squared as 𝑥 multiplied by 𝑥 is 𝑥 squared. Multiplying the outside terms gives us negative two 𝑥. Multiplying the inside terms gives us negative 𝑥 or negative one 𝑥. And multiplying the last terms gives us positive two as negative one multiplied by negative two is equal to positive two.

This can be simplified to 𝑥 squared minus three 𝑥 plus two. We now need to multiply this bracket by 𝑥 minus three. Multiplying 𝑥 squared by 𝑥 minus three gives us 𝑥 cubed minus three 𝑥 squared. Multiplying negative three 𝑥 by 𝑥 minus three gives us negative three 𝑥 squared plus nine 𝑥. And finally, multiplying two by 𝑥 minus three gives us two 𝑥 minus six.

Simplifying by grouping the like terms gives us 𝑥 cubed minus six 𝑥 squared plus eleven 𝑥 minus six. We can, therefore, rewrite the denominator as 𝑥 cubed minus six 𝑥 squared plus eleven 𝑥 minus six. Cross multiplying to eliminate the fractions gives us three multiplied by two 𝑥 minus four is equal to two multiplied by 𝑥 cubed minus six 𝑥 squared plus eleven 𝑥 minus six.

Expanding the brackets on both sides gives us six 𝑥 minus 12 is equal to two 𝑥 cubed minus 12𝑥 squared plus 22𝑥 minus 12. Subtracting six 𝑥 minus 12 from both sides of this equation gives us zero is equal to two 𝑥 cubed minus 12𝑥 squared plus 16𝑥 as 22 𝑥 minus six 𝑥 is equal to 16𝑥 and negative 12 minus negative 12 is equal to zero.

The three terms on the left-hand side of the equation have a common factor of two 𝑥. Therefore, we can factorize out two 𝑥. Two 𝑥 cubed divided by two 𝑥 is 𝑥 squared. Negative 12𝑥 squared divided by two 𝑥 is equal to negative six 𝑥. And 16𝑥 divided by two 𝑥 is equal to eight. Therefore, we are left with two 𝑥 multiplied by 𝑥 squared minus six 𝑥 plus eight.

We can then factorize the quadratic inside the bracket to 𝑥 minus two multiplied by 𝑥 minus four as the product of negative two and negative four is eight and the sum is negative six.

This equation gives us three possible solutions: two 𝑥 is equal to zero, 𝑥 minus two is equal to zero, or 𝑥 minus four is equal to zero. These three equations can be simplified to give us possible solutions of 𝑥 equals zero, 𝑥 equals two, and 𝑥 equals four.

Remember that in the question, we were told that 𝑥 could not equal one, two, or three as it would give us an undefined solution. This means that we can eliminate the solution 𝑥 equals two.

There are two possible solutions to the equation: 𝑥 equals zero or 𝑥 equals four. We could check both of these answers by substituting them into the left-hand side of the equation. In both cases, we would get an answer of two-thirds.