Question

Using cross multiplication method, solve \[3x + 5y = 25\] and $7x + 6y = 30$

Answer 1

Hint:To solve this problem we use cross multiplication method that is when the general equation is ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ by arranging this we can written it into
$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$ this form. where ${a_1},{b_1},{c_1}$ are coefficient of $x,y,$ and constant of equation first and similarly ${a_2},{b_2},{c_2}$ are coefficient of $x,y,$ and constant of equation second .

Complete step-by-step answer:
Given equations are \[3x + 5y - 25 = 0\] and $7x + 6y - 30 = 0$
So from first equation \[3x + 5y - 25 = 0\]
Compare it with general equation ${a_1}x + {b_1}y + {c_1} = 0$ we get
${a_1} = 3$ ; ${b_1} = 5$; and ${c_1} = - 25$
And similarly for equation second $7x + 6y - 30 = 0$
Compare it with general equation ${a_2}x + {b_2}y + {c_2} = 0$ we get
${a_2} = 7$; ${b_2} = 6$ and ${c_2} = - 30$
Now by using cross multiplication method
$\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$
Now put value of ${a_1},{b_1},{c_1}$ and ${a_2},{b_2},{c_2}$
$\dfrac{x}{{5 \times ( - 30) - 6 \times ( - 25)}} = \dfrac{y}{{\left( { - 25} \right) \times 7 - \left( { - 30} \right) \times 3}} = \dfrac{1}{{3 \times 6 - 7 \times 5}}$
Now solving each we get
$\dfrac{x}{{ - 150 + 150}} = \dfrac{y}{{ - 175 + 90}} = \dfrac{1}{{18 - 35}}$
From this
$\dfrac{x}{0} = \dfrac{y}{{ - 85}} = \dfrac{1}{{ - 17}}$
So now we solve a particular variable
$\dfrac{x}{0} = \dfrac{1}{{ - 17}}$
From this we get $x = 0$
And $\dfrac{y}{{ - 85}} = \dfrac{1}{{ - 17}}$
$y = \dfrac{{85}}{{17}}$
$y = 5$
So the solution of $x$ and $y$ are $0$ and $5$ respectively.

Note:We can solve this problem by balancing coefficients of any one of the given variables.
Balance the coefficient of $x$ by multiplying coefficient of given equation with each other (like coefficient of first equation multiplied with whole second equation and coefficient of second equation multiplied with whole first equation.
Given equation: \[3x + 5y = 25\] and $7x + 6y = 30$
So multiply equation \[3x + 5y = 25\] by $7$ and equation $7x + 6y = 30$ by $3$
So we get
\[(3x + 5y = 25) \times 7\]
$21x + 35y = 175$
And by solving second equation
 $(7x + 6y = 30) \times 3$
$21x + 18y = 90$
Now subtract both equation ( variable part is subtracted form variable part and constant is from constant)
$(21x + 35y) - (21x + 18y) = 175 - 90$
From this $17y = 85$
And $y = 5$
Now for value of $x$ put value of $y = 5$ in any one equation (original)
Choose the first equation \[3x + 5y = 25\] and substitute the above value.
\[3x + 5 \times 5 = 25\]
From this we get
$3x = 0$
And $x = 0$
So our answer is $x = 0$ and $y = 5$ .