Question

Which of the following is the solution set for the simultaneous equation?
\[\dfrac{2x – 1}{3} = \dfrac{y + 6}{5} = \dfrac{3x – y}{7}\]
A. \[x = 2\], \[y = - 1\]
B. \[x = 3\], \[y = - 1\]
C. \[x = - 1\], \[y = 2\]
D. \[x = 4\], \[y = 1\]

Answer 1

Hint: In this question, we need to find the value of \[x\] and \[y\] of the simultaneous equation. First, let us split the given set of simultaneous equations into two linear equations with two unknown variables . Then we can use the substitution method to find the value of \[x\] and \[y\]. By using the substitution method, first we can find the value of one variable from either equation, and then substitute that variable into the other equation to get another variable.

Complete step by step answer:
Given \[\dfrac{2x – 1}{3} = \dfrac{y + 6}{5} = \dfrac{3x – y}{7}\]. First, let us split the given set of simultaneous equations into two linear equations.By taking two equalities simultaneously,
We get,
\[\dfrac{2x – 1}{3} = \dfrac{y + 6}{5}\] and \[\dfrac{y + 6}{5} = \dfrac{3x – y}{7}\]
On cross multiplying,
We get,
\[5(2x – 1) = 3(y + 6)\] and \[7(y + 6) = 5(3x – y)\]
On simplifying,
We get,
\[\Rightarrow \ 10x – 5 = 3y + 18\] and \[7y + 42 = 15x – 5y\]
By making arrangements ,
We get,
\[10x – 3y = 18 + 5\] and \[15x – 5y – 7y = 42\]

On simplifying,
We get,
\[\Rightarrow \ 10x – 3y = 23\] and \[15x – 12y = 42\]
Thus we get two linear equations as \[10x – 3y = 23\] ••• (1) and \[15x – 12y = 42\] ••• (2)
Now let us consider equation (1) ,
\[10x – 3y = 23\]
On adding both sides by \[3y\] ,
We get,
\[10x = 23 + 3y\]
On dividing both sides by \[10\] ,
We get,
\[\Rightarrow \ x = \dfrac{23 + 3y}{10}\]

Now let us plug the value of \[x\] in either equation to find the value of \[y\].Now let us substitute the value of \[x\] in equation (1),
Equation (1) , \[15x – 12y = 42\]
By substituting \[x = \dfrac{23 + 3y}{10}\],
We get,
\[15\left( \dfrac{23 + 3y}{10} \right) – 12y = 42\]
On simplifying,
We get,
\[3\left( \dfrac{23 + 3y}{2} \right) – 12y = 42\]
On simplifying the numerator term,
We get,
\[\Rightarrow \dfrac{69 + 9y}{2} – 12y = 42\]

Now on taking LCM
We get,
\[\dfrac{\left( 69 + 9y \right) – 24y}{2} = 42\]
On cross multiplying,
We get,
\[69 + 9y – 24y = 42 \times 2\]
On simplifying,
We get
\[69 – 15y = 84\]
On subtracting both sides by \[69\] ,
We get,
\[\Rightarrow \ - 15y = 84 – 69\]
On simplifying,
We get,
\[- 15y = 15\]
On dividing by \[- 15\] ,
We get,
\[\Rightarrow \ y = - 1\]

Now let us plug the value of \[y\] in \[x = \dfrac{23 + 3y}{10}\] to get the value of \[x\].
On substituting \[y = - 1\] in \[x = \dfrac{23 + 3y}{10}\]
\[\Rightarrow \ x = \dfrac{23 + 3\left( - 1 \right)}{10}\]
On simplifying,
We get,
\[\Rightarrow \ x = \dfrac{23 – 3}{10}\]
On further simplifying,
We get,
\[\Rightarrow \ x = \dfrac{20}{10}\]
On dividing,
We get ,
\[x = 2\]
Thus we get the value of \[x\] and \[y\] as \[2\] and \[- 1\] respectively.Hence, the solution set for the simultaneous equation \[\dfrac{2x – 1}{3} = \dfrac{y + 6}{5} = \dfrac{3x – y}{7}\] is \[x = 2\] and \[y = - 1\] .

Therefore, option A is the correct answer.

Note: We should not get confused in choosing the equation that we want to work with , it doesn’t matter which equation we are choosing , but we can choose the one which makes our calculations easy. In order to solve a pair of simultaneous linear equations containing two variables, we usually used to prefer an algebraic method.