Question

Solve the system of equations \[2x + 3y = 17,3x - 2y = 6\] by the method of cross multiplication.
A.\[x = 4,y = 3\]
B.\[x = 5,y = 3\]
C.\[x = 4,y = 4\]
D.None of these

Answer 1

Hint: Here we will use the basic concept of the cross multiplication method to solve the equation. First, we will convert the given equation in the standard equation. Then we will apply the basic formula and solve it further to find the value of \[x\] and \[y\].

Formula used:
For the equations \[{a_1}x + {b_1}y + {c_1} = 0,{a_2}x + {b_2}y + {c_2} = 0\] by the cross multiplication method we get \[x = \dfrac{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\] and \[y = \dfrac{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\]

Complete step-by-step answer:
Given equations are \[2x + 3y = 17,3x - 2y = 6\].
Now we will use the basic formula of the method of cross multiplication.
We will convert the given equation in the standard form \[{a_1}x + {b_1}y + {c_1} = 0,{a_2}x + {b_2}y + {c_2} = 0\]. Therefore, we get
\[\begin{array}{l}2x + 3y - 17 = 0\\3x - 2y - 6 = 0\end{array}\]
We will substitute the values in the formula to get the values of \[x\] and \[y\].
First, we will find the value of \[x\] using the formula \[x = \dfrac{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\]. Therefore, we get
\[x = \dfrac{{\left( {3\left( { - 6} \right) - \left( { - 2} \right)\left( { - 17} \right)} \right)}}{{\left( {2\left( { - 2} \right) - 3 \times 3} \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow x = \dfrac{{\left( { - 18 - 34} \right)}}{{\left( { - 4 - 9} \right)}}\]
Adding the terms, we get
\[ \Rightarrow x = \dfrac{{ - 52}}{{ - 13}}\]
Dividing the terms, we get
\[ \Rightarrow x = 4\]
Now we will find the value of \[y\] using the formula \[y = \dfrac{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\]. Therefore, we get
\[y = \dfrac{{\left( {\left( { - 17} \right)3 - \left( { - 6} \right)2} \right)}}{{\left( {2\left( { - 2} \right) - 3 \times 3} \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow y = \dfrac{{\left( { - 51 + 12} \right)}}{{\left( { - 4 - 9} \right)}}\]
Adding the terms, we get
\[ \Rightarrow y = \dfrac{{ - 39}}{{ - 13}}\]
Dividing the terms, we get
\[ \Rightarrow y = 3\]
Hence the value of \[x = 4\] and \[y = 3\].
So, option A is the correct option.

Note: We should know the basic formula of the cross multiplication method. A cross multiplication method is generally applied to the equation having two variables. It cannot be applied to the equation with a simple variable like linear or quadratic equation. A linear equation is the equation in which the highest exponent of the variable x is one. A quadratic equation is an equation in which the highest exponent of the variable x is two and a quadratic equation has only two roots.