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
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
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.