Question

Find the value of x and y using cross multiplication method:
\[2x+y=13\] and \[x+y=8\].
A. (5, 3)
B. (5, -3)
C. (-5, -3)
D. (-5, 3)

Answer 1

Hint:Consider two linear equations applying the cross product where the coefficient of x, y and constant terms in the equation are interrelated. Take this relation into formula.

Complete step-by-step answer:
We have been given two equations \[2x+y=13.....(1)\]
\[x+y=8....(2)\]
Now let us look into the formula for cross multiplication and its use in solving two simultaneous equations can be presented as,
\[\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0......(3) \\
 & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0.....(4) \\
\end{align}\]
Thus we can draw it as,

Here z=1
Thus from this we can write as,
\[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]
From this we can write,
\[\begin{align}
  & \dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
 & \Rightarrow x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}.....(5) \\
\end{align}\]
\[\begin{align}
  & \dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\
 & \therefore y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}....(6) \\
\end{align}\]
Thus from equation (1) , \[2x+y-13=0\], compare it with equation (3).
\[{{a}_{1}}=2,{{b}_{1}}=1,{{c}_{1}}=-13\]
Now from equation (2), \[x+y=8\], compare it with equation (4).
\[{{a}_{2}}=1,{{b}_{2}}=1,{{c}_{2}}=-8\]
Now apply these values in equation (5) and (6).

Here z=1
\[\begin{align}
  & x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}=\dfrac{1\times (-8)-1\times (-13)}{2\times 1-1\times 1}=\dfrac{-8+13}{2-1}=\dfrac{5}{1}=5 \\
 & y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}=\dfrac{(-13\times 1)-(-8\times 2)}{(2\times 1)-(1\times 1)}=\dfrac{-13+16}{2-1}=3 \\
\end{align}\]
Thus we got x = 5 and y = 3, i.e. the value we got by cross multiplication is (5, 3).
Option A is the correct answer.

Note: If the value of x or y is zero, that is \[\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)=0\] or \[\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)=0\], it is not proper way to express in the formula for cross multiplication, because the denominator of a fraction can never be zero. For two simultaneous equations, cross multiplication is an important concept.