Question

Solve the following pair of equations:
$\dfrac{x}{8} - \dfrac{y}{7} = \dfrac{{17}}{{28}};\dfrac{x}{7} - \dfrac{y}{8} = 1$
A.$x = 5,y = 8$
B.$x = 19,y = 8$
C.$x = 12,y = 8$
D.$x = 14,y = 8$

Answer 1

Hint: Take the given two expression and simplify the terms in the form of the fraction in the terms of the simple terms by using the cross-multiplication and LCM (least common multiple) and make one of the variables in both the equations with the same coefficient and use elimination method to find the values for “x” and “y”.

Complete step-by-step answer:
Take the given expressions:
$\dfrac{x}{8} - \dfrac{y}{7} = \dfrac{{17}}{{28}};\dfrac{x}{7} - \dfrac{y}{8} = 1$
Use the LCM (least common multiple) for the above expressions and simplify using the cross-multiplication. In cross-multiplication the denominator of one side is multiplied with the numerator of the opposite side.
$
  \dfrac{x}{8} - \dfrac{y}{7} = \dfrac{{17}}{{28}} \\
  \dfrac{{7x - 8y}}{{56}} = \dfrac{{17}}{{28}} \\
  7x - 8y = \dfrac{{17}}{{28}} \times 56 \\
 $
Find the factors for the term on the right hand side of the equation –
$7x - 8y = \dfrac{{17 \times 28 \times 2}}{{28}}$
Common factors from the numerator and the denominator cancel each other.
$7x - 8y = 34$ (A)
Similarly, for the second equation –
$\dfrac{x}{7} - \dfrac{y}{8} = 1$
Take LCM for the above expression –
$\dfrac{{8x - 7y}}{{56}} = 1$
Perform Cross-multiplication in the above expression –
$8x - 7y = 56$ (B)
Multiply equation (A) by 8 and the equation (B) 7 by
$56x - 64y = 272$ …. (C)
$56x - 49y = 392$ … (D)
Subtract equation (C) from (D)
$(56x - 49y) - (56x - 64y) = 392 - 272$
When there is a negative sign outside the bracket then the sign of the terms inside the bracket changes. Positive term changes to negative and vice-versa.
$56x - 49y - 56x + 64y = 120$
Like terms with the same value and opposite sign cancels each other.
$ - 49y + 64y = 120$
Simplify the above expression finding the difference of the terms-
$15y = 120$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$y = \dfrac{{120}}{{15}}$
Common multiples from the numerator and the denominator cancels each other.
$ \Rightarrow y = 8$ …. (E)
Place the above value in the equation (B) $8x - 7y = 56$
$8x - 7(8) = 56$
Make the required term the subject –
$
  8x - 56 = 56 \\
  8x = 56 + 56 \\
  8x = 112 \;
 $
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$x = \dfrac{{112}}{8}$
Simplify –
$x = 14$ …. (F)
Hence, from the given multiple choices – the option D is the correct answer.
So, the correct answer is “Option D”.

Note: Be careful about the sign convention while simplification. Remember when there is negative sign outside the bracket then the sign of the terms inside the bracket also changes when brackets are opened. Negative terms are changed to positive and vice-versa.
We needed to find the value of two unknowns hence minimum two or more variables are required.