Question
Introduction
Somebody
said, “Dear Algebra, please stop asking me about your x. She is not coming back. Don’t ask me
y!”
Here we have
here linear equations that appear to be more tricky than the usual fanfare. This is because the coefficients of the
unknowns x and y are fractions. In school, students learn to solve these
using the method of substitution and
the method of elimination. They tend to prefer to do it by the former
method, because psychologically it seems easier to accept. However, the latter matter is generally more
effective and yields a shorter solution. This question is like a fly trap for those who
prefer the method of substitution. You
make either x or y the subject from one of the equations, and then
substitute that into the other equation. This yields a complicated algebraic fractions
within algebraic fractions. The more
complicated your solution is, the higher chance there is for making careless
mistakes. It is a good idea that
students get out of their comfort zones and adopt a new skill. So how do we do it?
The smart tactic
First, we
need to clear away the fractions by multiplying through with the LCM of the
denominators appearing in each equation. For the first equation, LCM(4, 8) = 8. Since
4 is a factor of 8, 8 is like a giant that absorbs the number 4. OK,
so we multiply the first equation through by
8. For the second equation, we
multiply every term by LCM (3, 2) = 6
Solution
Remarks
As you can
see, equations [3] and [4] both contain – 3y and this can be eliminated via subtraction. So the value of x comes out easily. Now once
x is known, one can find the y!
For another example of simultaneous equations, please refer
to this article.
H04. Look for
pattern(s)
H05. Work
backwards
H10. Simplify
the problem
H11. Solve part
of the problem
H13* Use
Equation / write a Mathematical Sentence
Suitable Levels
* Lower Secondary Mathematics (Secondary 2)
* GCE ‘O’ Level “Elementary” Mathematics (revision)
* other syllabuses that involve algebra,
linear simultaneous equations
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