Question

Solve the following pair of linear equations by the substitution method.
1. $x + y = 14$; $x - y = 4$
2. $s - t = 3$; $\dfrac{s}{3} + \dfrac{t}{2} = 6$
3. $3x - y = 3$; $9x - 3y = 9$
4. $0.2x + 0.3y = 1.3$; $0.4x + 0.5y = 2.3$
5. $\sqrt 2 x + \sqrt 3 y = 0$; $\sqrt 3 x - \sqrt 8 y = 0$
6. $\dfrac{{3x}}{2} - \dfrac{{5y}}{3} = - 2$; $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{{13}}{6}$

Answer 1

Hint: We are given linear equations in two variables and we need to solve them by substitution method. In the substitution method, we express one of the variables in terms of the other variable from either of the two equations and then this expression is put in the other equation to obtain an equation in one variable.

Complete step-by-step solution:
1. $x + y = 14$; $x - y = 4$
The given system of equations is
$x + y = 14.....\left( i \right)$
$x - y = 4.....\left( {ii} \right)$
From $\left( {ii} \right)$, we get
$ \Rightarrow x = y + 4$
Substituting $x = y + 4$ in $\left( i \right)$, we get
$x + y = 14.....\left( i \right)$
$ \Rightarrow y + 4 + y = 14$
On shifting $4$ to RHS, we get
$ \Rightarrow y + y = 14 - 4$
On addition and subtraction of like terms, we get
$ \Rightarrow 2y = 10$
On division, we get
$ \Rightarrow y = \dfrac{{10}}{2}$
$ \Rightarrow y = 5$
Substitute $y = 5$ in $\left( {ii} \right)$
$x - y = 4.....\left( {ii} \right)$
$ \Rightarrow x - 5 = 4$
$ \Rightarrow x = 9$
Hence, the solution of the given system of equations is $x = 9$, $y = 5$.

2. $s - t = 3$; $\dfrac{s}{3} + \dfrac{t}{2} = 6$
The given system of equations is
$s - t = 3.....\left( i \right)$
$\dfrac{s}{3} + \dfrac{t}{2} = 6.....\left( {ii} \right)$
From $\left( i \right)$, we get
\[ \Rightarrow s = 3 + t\]
Substituting \[s = 3 + t\] in $\left( {ii} \right)$, we get
$\dfrac{s}{3} + \dfrac{t}{2} = 6.....\left( {ii} \right)$
$ \Rightarrow \dfrac{{3 + t}}{3} + \dfrac{t}{2} = 6$
On taking LCM on LHS, we get
$ \Rightarrow \dfrac{{2\left( {3 + t} \right) + 3t}}{6} = 6$
On cross-multiplication, we get
$ \Rightarrow 2\left( {3 + t} \right) + 3t = 6 \times 6$
After solving bracket, we get
$ \Rightarrow 6 + 2t + 3t = 36$
$ \Rightarrow 5t = 36 - 6$
$ \Rightarrow 5t = 30$
$ \Rightarrow t = \dfrac{{30}}{5}$
On division, we get
$ \Rightarrow t = 6$
Putting $t = 6$ in $\left( i \right)$
$s - t = 3.....\left( i \right)$
\[ \Rightarrow s - 6 = 3\]
\[ \Rightarrow s = 9\]
Hence, the solution of the given system of equations is $s = 9$, $t = 6$.

3. $3x - y = 3$; $9x - 3y = 9$
The given system of equations is
$3x - y = 3.....\left( i \right)$
$9x - 3y = 9.....\left( {ii} \right)$
From $\left( i \right)$
$3x - y = 3.....\left( i \right)$
On shifting $y$ to RHS, we get
$ \Rightarrow 3x = y + 3$
$ \Rightarrow x = \dfrac{{y + 3}}{3}$
Substituting $x = \dfrac{{y + 3}}{3}$ in $\left( {ii} \right)$, we get
$9x - 3y = 9.....\left( {ii} \right)$
$ \Rightarrow 9\left( {\dfrac{{y + 3}}{3}} \right) - 3y = 9$
On cancelling out, we get
$ \Rightarrow 3\left( {y + 3} \right) - 3y = 9$
After solving bracket, we get
$ \Rightarrow 3y + 9 - 3y = 9$
$ \Rightarrow 9 = 9$
The statement is true for all values of $x$.
So, there are infinitely many solutions.

4. $0.2x + 0.3y = 1.3$; $0.4x + 0.5y = 2.3$
The given system of equations is
$0.2x + 0.3y = 1.3.....\left( i \right)$
$0.4x + 0.5y = 2.3.....\left( {ii} \right)$
Let us multiply equation $\left( i \right)$ by $10$ on both sides
$ \Rightarrow \left( {0.2x + 0.3y} \right) \times 10 = 1.3 \times 10$
$ \Rightarrow 2x + 3y = 13$
On shifting $3y$ to RHS, we get
$ \Rightarrow 2x = 13 - 3y$
$ \Rightarrow x = \dfrac{{13 - 3y}}{2}$
Substituting $x = \dfrac{{13 - 3y}}{2}$ in $\left( {ii} \right)$, we get
$ \Rightarrow 0.4\left( {\dfrac{{13 - 3y}}{2}} \right) + 0.5y = 2.3$
Multiplying both sides by $10$
$ \Rightarrow 10 \times 0.4\left( {\dfrac{{13 - 3y}}{2}} \right) + 0.5y = 2.3 \times 10$
$ \Rightarrow 4\left( {\dfrac{{13 - 3y}}{2}} \right) + 0.5y = 23$
On cancelling out, we get
$ \Rightarrow 2\left( {13 - 3y} \right) + 5y = 23$
On multiplication, we get
$ \Rightarrow 26 - 6y + 5y = 23$
$ \Rightarrow 26 - y = 23$
On subtraction, we get
$ \Rightarrow y = 26 - 23$
$ \Rightarrow y = 3$
Substitute value of $y = 3$ in equation $\left( i \right)$
$0.2x + 0.3y = 1.3.....\left( i \right)$
$ \Rightarrow 0.2x + 0.3 \times 3 = 1.3$
On multiplication, we get
$ \Rightarrow 0.2x + 0.9 = 1.3$
Shift $0.9$ to RHS
$ \Rightarrow 0.2x = 1.3 - 0.9$
$ \Rightarrow 0.2x = 0.4$
$ \Rightarrow x = \dfrac{{0.4}}{{0.2}}$
On division, we get
$ \Rightarrow x = 2$
Hence, the solution of the given system of equations is $x = 2$, $y = 3$.

5. $\sqrt 2 x + \sqrt 3 y = 0$; $\sqrt 3 x - \sqrt 8 y = 0$
The given system of equations is
$\sqrt 2 x + \sqrt 3 y = 0.....\left( i \right)$
$\sqrt 3 x - \sqrt 8 y = 0.....\left( {ii} \right)$
From $\left( i \right)$
$ \Rightarrow \sqrt 2 x + \sqrt 3 y = 0$
On shifting $\sqrt 3 y$ to RHS, we get
\[ \Rightarrow \sqrt 2 x = - \sqrt 3 y\]
\[ \Rightarrow x = \dfrac{{ - \sqrt 3 y}}{{\sqrt 2 }}\]
Substituting \[x = \dfrac{{ - \sqrt 3 y}}{{\sqrt 2 }}\] in $\left( {ii} \right)$, we get
$\sqrt 3 x - \sqrt 8 y = 0.....\left( {ii} \right)$
$ \Rightarrow \sqrt 3 \left( {\dfrac{{ - \sqrt 3 y}}{{\sqrt 2 }}} \right) - \sqrt 8 y = 0$
On multiplication, we get
$ \Rightarrow \dfrac{{ - 3y}}{{\sqrt 2 }} - \sqrt 8 y = 0$
Take LCM on left-hand side
$ \Rightarrow \dfrac{{ - 3y - \sqrt 8 y \times \sqrt 2 }}{{\sqrt 2 }} = 0$
On cross-multiplication, we get
$ \Rightarrow - 3y - \sqrt {16} y = 0$
We know that the square root of $16$ is $4$.
$ \Rightarrow - 3y - 4y = 0$
On solving negative terms, we get
$ \Rightarrow - 7y = 0$
$ \Rightarrow y = \dfrac{0}{{ - 7}}$
On division, we get
$ \Rightarrow y = 0$
Substitute value of $y = 0$ in equation $\left( i \right)$
$\sqrt 2 x + \sqrt 3 y = 0.....\left( i \right)$
$ \Rightarrow \sqrt 2 x + \sqrt 3 \times 0 = 0$
Any number multiplied by zero becomes zero. So, we get
$ \Rightarrow \sqrt 2 x = 0$
\[ \Rightarrow x = \dfrac{0}{{\sqrt 2 }}\]
On division, we get
$ \Rightarrow x = 0$
Hence, the solution of the given system of equations is $x = 0$, $y = 0$.

6. $\dfrac{{3x}}{2} - \dfrac{{5y}}{3} = - 2$; $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{{13}}{6}$
Let us first remove fractions from both equations.
$\dfrac{{3x}}{2} - \dfrac{{5y}}{3} = - 2$
Multiplying both sides by $6$
$6 \times \dfrac{{3x}}{2} - 6 \times \dfrac{{5y}}{3} = 6 \times - 2$
$9x - 10y = - 12....\left( i \right)$
Another equation,
$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{{13}}{6}$
Multiplying both sides by $6$
$6 \times \dfrac{x}{3} + 6 \times \dfrac{y}{2} = 6 \times \dfrac{{13}}{6}$
$2x + 3y = 13.....\left( {ii} \right)$
Hence, our equations are
$9x - 10y = - 12....\left( i \right)$
$2x + 3y = 13.....\left( {ii} \right)$
From $\left( i \right)$,
$ \Rightarrow 9x - 10y = - 12$
Shift $ - 10y$ to RHS.
$ \Rightarrow 9x = - 12 + 10y$
$ \Rightarrow x = \dfrac{{ - 12 + 10y}}{9}$
Substituting value of $x = \dfrac{{ - 12 + 10y}}{9}$ in $\left( {ii} \right)$
$ \Rightarrow 2\left( {\dfrac{{ - 12 + 10y}}{9}} \right) + 3y = 13$
Multiply both sides by $9$
$ \Rightarrow 9 \times 2\left( {\dfrac{{ - 12 + 10y}}{9}} \right) + 9 \times 3y = 9 \times 13$
On cancelling out, we get
$ \Rightarrow 2\left( { - 12 + 10y} \right) + 27y = 9 \times 13$
On solving bracket, we get
$ \Rightarrow - 24 + 20y + 27y = 117$
Shift $ - 24$ to RHS
$ \Rightarrow 20y + 27y = 117 + 24$
On addition, we get
$ \Rightarrow 47y = 141$
$ \Rightarrow y = \dfrac{{141}}{{47}}$
On division, we get
$ \Rightarrow y = 3$
Substitute value of $y = 3$ in equation $\left( {ii} \right)$
$ \Rightarrow 2x + 3y = 13.....\left( {ii} \right)$
On multiplication, we get
$ \Rightarrow 2x + 3 \times 3 = 13$
$ \Rightarrow 2x + 9 = 13$
Shift $9$ to RHS
$ \Rightarrow 2x = 13 - 9$
$ \Rightarrow 2x = 4$
$ \Rightarrow x = \dfrac{4}{2}$
On division, we get
$ \Rightarrow x = 2$
Hence, the solution of the given system of equations is $x = 2$, $y = 3$.

Note: Here, in the given question, a particular method is mentioned to solve the linear equations. But there are many other methods by which we can solve linear equations. The most commonly used algebraic methods of solving linear equations in two variables are: method of elimination by substitution, method of elimination by equating the coefficients and method of cross-multiplication.