Question

Solve the equations using cross multiplication method:
$x-3y=8$ and $x-9y=14$
a) $x=-5\text{ and }y=1$
b) $x=5\text{ and }y=1$
c) $x=5\text{ and }y=-1$
d) $x=-5\text{ and }y=-1$

Answer 1

Hint: We get the values for ${{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}},{{c}_{2}}$ from the given two equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$. We check whether ${{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}\ne 0$ . Then by cross multiplication method, we find $x$ and $y$ using the formula
$\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}$ and hence \[x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}\] and $y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}$

Complete step-by-step solution:
Given equations are $x-3y=8$ and $x-9y=14$ . Now bringing all terms to the left hand side of the equations, we get $x-3y-8=0$ and $x-9y-14=0$ . By comparing with ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$, we get ${{a}_{1}}=1$, ${{b}_{1}}=-3$, ${{c}_{1}}=-8$, ${{a}_{2}}=1$,${{b}_{2}}=-9$ and ${{c}_{2}}=-14$.
Here ${{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}=(-9)(1)-(-3)(1)$
$\Rightarrow {{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}=-9+3$
$\Rightarrow {{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}=-6$ and hence ${{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}\ne 0$
By the cross multiplication method,
$\Rightarrow \dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}$
Now applying the above formula for the values of ${{a}_{1}},{{b}_{1}},{{c}_{1}},{{a}_{2}},{{b}_{2}},{{c}_{2}}$ we get $\Rightarrow \dfrac{x}{(-3)(-14)-(-9)(-8)}=\dfrac{y}{(-8)(1)-(-14)(1)}=\dfrac{1}{(-9)(1)-(-3)(1)}$
On multiplying the values in each of the denominator, we get
$\Rightarrow \dfrac{x}{42-72}=\dfrac{y}{-8+14}=\dfrac{1}{-9+3}$
Now adding the terms in each denominator, we get
$\Rightarrow \dfrac{x}{-30}=\dfrac{y}{6}=\dfrac{1}{-6}.............(i)$
Now to find x, we equate the “x” part with the constant term from the above expression (i).
$\Rightarrow \dfrac{x}{-30}=\dfrac{1}{-6}$
By using the formula \[x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}\] , we get $x=\dfrac{-30}{-6}$ , which on simplification we get $x=5$
Now, to find y, we equate the “y” part with the constant term from the above expression (i).
$\Rightarrow \dfrac{y}{6}=\dfrac{1}{-6}$
By using the formula $y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}}}$, we get $y=\dfrac{6}{-6}$ , which on simplification we get $y=-1$
Hence the correct answer is c) $x=5\text{ and }y=-1$

Note: We can verify the correctness of the answer by substituting the values of x and y in the given two equations.
If $x=5\text{ and }y=-1$ , the first equation becomes $x-3y=5-(3)(-1)=8$ and the second equation becomes $x-9y=5-(9)(-1)=14$
Thus our obtained answer is the correct answer.