Question

Solve \[3\left( {2x + y} \right) = 7xy\] and \[3\left( {x + 3y} \right) = 11xy\] using elimination method.

Answer 1

Hint: Here, we need to solve the given equations using elimination method. We will form two linear equations in terms of two distinct variables. We will solve the two linear equations using elimination methods to find the values \[x\] and \[y\].

Complete step by step solution:
The given equations \[3\left( {2x + y} \right) = 7xy\] and \[3\left( {x + 3y} \right) = 11xy\] are linear equations in two variables.
Multiplying the terms in the equation \[3\left( {2x + y} \right) = 7xy\] using distributive law of multiplication, we get
\[ \Rightarrow 6x + 3y = 7xy\]
Dividing both sides by \[xy\], we get
\[ \Rightarrow \dfrac{{6x + 3y}}{{xy}} = \dfrac{{7xy}}{{xy}}\]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{{6x}}{{xy}} + \dfrac{{3y}}{{xy}} = \dfrac{{7xy}}{{xy}}\]
\[ \Rightarrow \dfrac{6}{y} + \dfrac{3}{x} = 7\]…………….\[\left( 1 \right)\]
Multiplying the terms in the equation \[3\left( {x + 3y} \right) = 11xy\] using distributive law of multiplication, we get
\[ \Rightarrow 3x + 9y = 11xy\]
Dividing both sides of the equation by \[xy\], we get
\[ \Rightarrow \dfrac{{3x + 9y}}{{xy}} = \dfrac{{11xy}}{{xy}}\]
Simplifying the expression, we get
\[ \Rightarrow \dfrac{{3x}}{{xy}} + \dfrac{{9y}}{{xy}} = \dfrac{{11xy}}{{xy}}\]
\[ \Rightarrow \dfrac{3}{y} + \dfrac{9}{x} = 11\]……………..\[\left( 2 \right)\]
Let \[p = \dfrac{1}{x}\] and \[q = \dfrac{1}{y}\].
Rewriting the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[ \Rightarrow 6q + 3p = 7\]………….\[\left( 3 \right)\]
\[ \Rightarrow 3q + 9p = 11\]……….\[\left( 4 \right)\]
We will use elimination method to solve these two linear equations.
Multiplying both sides of equation \[\left( 4 \right)\] by 2 using distributive law of multiplication, we get
\[ \Rightarrow \left( {3q + 9p} \right) \times 2 = 11 \times 2\]
\[ \Rightarrow 6q + 18p = 22\]…………………..\[\left( 5 \right)\]
Subtracting both sides of equation \[\left( 3 \right)\] from equation \[\left( 5 \right)\], we get
\[\begin{array}{l}6q + 18p = 22\\\underline {6q + {\rm{ }}3p = {\rm{ }}7} \\{\rm{ }}15p{\rm{ = 15}}\end{array}\]
Dividing both sides by 15, we get
\[\begin{array}{l} \Rightarrow \dfrac{{15p}}{{15}} = \dfrac{{15}}{{15}}\\ \Rightarrow p = 1\end{array}\]
Substituting \[p = 1\] in the equation \[6q + 3p = 7\], we get
\[ \Rightarrow 6q + 3\left( 1 \right) = 7\]
Multiplying the terms in the expression, we get
\[ \Rightarrow 6q + 3 = 7\]
Subtracting 3 from both sides, we get
\[ \Rightarrow 6q = 4\]
Dividing both sides by 6, we get
\[ \Rightarrow q = \dfrac{4}{6}\]
Simplifying the expression, we get
\[ \Rightarrow q = \dfrac{2}{3}\]
Substituting \[p = 1\] and \[q = \dfrac{2}{3}\] in the equations \[p = \dfrac{1}{x}\] and \[q = \dfrac{1}{y}\], we get
\[ \Rightarrow 1 = \dfrac{1}{x}\] and \[\dfrac{2}{3} = \dfrac{1}{y}\]
Simplifying the two equations, we get
\[x = 1\] and \[y = \dfrac{3}{2}\]
\[\therefore \] We get the values of \[x\] and \[y\] as 1 and \[\dfrac{3}{2}\].

Note: We have formed two linear equations in two variables and solved them using elimination method to find the values of \[x\] and \[y\]. A linear equation in two variables is an equation of the form \[ax + by + c = 0\], where \[a\] and \[b\]are not equal to 0. For example, \[2x - 7y = 4\] is a linear equation in two variables.
We have used the distributive law of multiplication to multiply the terms in the solution. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].