Question

Solve the given pair of equations:
$
  0.4x - 1.5y = 6.5 \\
  0.3x + 0.2y = 0.9 \\
$

Answer 1

Hint: We first multiply the pair of equations by 10 to avoid decimal in the equations. Then, we will use a cross-multiplication method to solve the equations. If $ax + by + c = 0$ and $px + qy + r = 0$, then $\dfrac{x}{{br - qc}} = - \dfrac{y}{{ar - cp}} = \dfrac{1}{{aq - bp}}$ . Substitute the value sand solve for the value of $x$ and $y$

Complete step-by-step answer:
We are given a pair of equations in two variables.
$
  0.4x - 1.5y = 6.5 \\
  0.3x + 0.2y = 0.9 \\
$
We have to find the find of $x$ and $y$ that satisfies both the above equations.
Let us remove the decimals by multiplying both the equations by 10.
$
  4x - 15y = 65 \\
  3x + 2y = 9 \\
$
We will solve the equations by cross multiplication method.
On rearranging the equations, we will get,
$
  4x - 15y - 65 = 0 \\
  3x + 2y - 9 = 0 \\
$
If $ax + by + c = 0$ and $px + qy + r = 0$, then \[\dfrac{x}{{br - qc}} = - \dfrac{y}{{ar - cp}} = \dfrac{1}{{aq - bp}}\]
Then, from the given pair of equations, we have $\dfrac{x}{{\left( { - 15} \right)\left( { - 9} \right) - \left( { - 65} \right)\left( 2 \right)}} = - \dfrac{y}{{\left( 4 \right)\left( { - 9} \right) - \left( { - 65} \right)\left( 3 \right)}} = \dfrac{1}{{\left( 4 \right)\left( 2 \right) - \left( { - 15} \right)\left( 3 \right)}}$
On solving it further, we will get,
$
  \dfrac{x}{{135 + 130}} = - \dfrac{y}{{ - 36 + 195}} = \dfrac{1}{{8 + 45}} \\
  \dfrac{x}{{265}} = - \dfrac{y}{{159}} = \dfrac{1}{{53}} \\
 $
Since, all are equal to each other, we will get,
$
  \dfrac{x}{{265}} = \dfrac{1}{{53}} \\
  x = \dfrac{{265}}{{53}} = 5 \\
$
$
   - \dfrac{y}{{159}} = \dfrac{1}{{53}} \\
  y = - \dfrac{{159}}{{53}} = - 3 \\
 $
Hence, the value of $x$ is 5 and value of $y$ is $ - 3$

Note: We can also calculate the values using elimination, substitution or graphical method to solve these types of questions. Also, if $ax + by + c = 0$ and $px + qy + r = 0$ are two equations and $\dfrac{a}{p} \ne \dfrac{b}{q}$, then we get a unique solution of the equations.