Question

Solve the following pairs of equations by reducing them to a pair of linear equations-
$
  \left( i \right)\dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2;\dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6} \\
  \left( {ii} \right)\dfrac{2}{{\sqrt x }} + \dfrac{3}{{\sqrt y }} = 2;\dfrac{4}{{\sqrt x }} - \dfrac{9}{{\sqrt y }} = - 1 \\
  \left( {iii} \right)\dfrac{4}{x} + 3y = 14;\dfrac{3}{x} - 4y = 23 \\
  \left( {iv} \right)\dfrac{5}{{x - 1}} + \dfrac{1}{{y - 2}} = 2;\dfrac{6}{{x - 1}} - \dfrac{3}{{y - 2}} = 1 \\
  \left( v \right)\dfrac{{7x - 2y}}{{xy}} = 5;\dfrac{{8x + 7y}}{{xy}} = 15 \\
  \left( {vi} \right)6x + 3y = 6xy;2x + 4y = 5xy \\
  \left( {vii} \right)\dfrac{{10}}{{x + y}} + \dfrac{2}{{x - y}} = 4;\dfrac{{15}}{{x + y}} - \dfrac{5}{{x - y}} = - 2 \\
  \left( {viii} \right)\dfrac{1}{{3x + y}} + \dfrac{1}{{3x - y}} = \dfrac{3}{4};\dfrac{1}{{2\left( {3x + y} \right)}} - \dfrac{1}{{2\left( {3x - y} \right)}} = \dfrac{{ - 1}}{8} \\
 $

Answer 1

Hint-Here we will proceed by selecting one of the methods to solve these equations. In this question, we will use a substitution method to solve the equations and get the required values of the equations.

Complete step-by-step solution -
$\left( i \right)\dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2;\dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}$
Solution - Let us assume $\dfrac{1}{x} = u,\dfrac{1}{y} = v$ in equation 1 and 2.
$
  \dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2...........(1) \\
  \dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}............(2) \\
 $
So the equations become-
Using equation 1, we get
$\dfrac{1}{2}u + \dfrac{1}{3}v = 2$
Or $\dfrac{{3u + 2v}}{{2 \times 3}} = 2$
Or 3u + 2v = 12 ………………… (3)
Using equation 2, we get
$\dfrac{1}{3}u + \dfrac{1}{2}v = \dfrac{{13}}{6}$
Or $\dfrac{{2u + 3v}}{{2 \times 3}} = \dfrac{{13}}{6}$
 Or 2u + 3v = 13 ……………… (4)
Now we will apply substitution method to solve these linear equations-
Firstly, we will take equation 3,
2u + 2v = 12
Or $u = \dfrac{{12 - 2v}}{3}$
Putting value of u in equation 4,
2u + 3v = 13
Or $2\left( {\dfrac{{12 - 2v}}{3}} \right) + 3v = 13$
Multiplying both sides by 3,
$3 \times 2\left( {\dfrac{{12 - 2v}}{3}} \right) + 3 \times 3v = 3 \times 13$
Or 2(12 – 2v) + 9v = 39
Or 24 – 4v + 9v = 39
Or -4v + 9v = 39 – 24
Or 5v = 15
Or v = 3
Now we will put the value of v=3 in equation 3,
3u + 2v = 12
Or 3u + 2(3) = 12
Or 3u + 6 = 12
Or 3u = 12 – 6
Or 3u = 6
Or u = 2
Hence we get v = 3 and u = 2
But we have to find x and y,
We know that
$u = \dfrac{1}{x}$
Or $x = \dfrac{1}{2}$
Similarly we assumed$v = \dfrac{1}{y}$
or $3 = \dfrac{1}{y}$
or $y = \dfrac{1}{3}$
So $x = \dfrac{1}{2}$ and $y = \dfrac{1}{3}$is the solution of the given equation i.e. $\dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2;\dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}$.

$\left( {ii} \right)\dfrac{2}{{\sqrt x }} + \dfrac{3}{{\sqrt y }} = 2;\dfrac{4}{{\sqrt x }} - \dfrac{9}{{\sqrt y }} = - 1$

Solution – Let us assume $\dfrac{1}{{\sqrt x }} = u;\dfrac{1}{{\sqrt y }} = v$
Now put the value of u and v in equations 1 and 2 i.e. $\begin{gathered}
  \dfrac{2}{{\sqrt x }} + \dfrac{3}{{\sqrt y }} = 2.........(1) \\
  \dfrac{4}{{\sqrt x }} - \dfrac{9}{{\sqrt y }} = - 1..........(2) \\
\end{gathered} $
So our equations will become-
2u + 3v = 2………. (3)
4u – 9v = -1 ………… (4)
Now by using equation 3 we will get the value of u-
2u + 3v = 2
Or 2u = 2 – 3v
Or $u = \dfrac{{2 - 3v}}{2}$
We will put the value of u in equation 4-
2(2-3v) – 9v = -1
Or 4 - 6v – 9v = -1
Or -15v = -5
Or $v = \dfrac{1}{3}$
Also we will substitute the value of v in equation 3-
2u + 3v = 2
Or $2u + 3\left( {\dfrac{1}{3}} \right) = 2$
Or 2u + 1 = 2
Or 2u = 1
Or $u = \dfrac{1}{2}$
Hence $u = \dfrac{1}{2}$and $v = \dfrac{1}{3}$
But we need to find the value of x and y-
$\dfrac{1}{{\sqrt x }} = u$(calculated)
Or $\dfrac{1}{2} = \dfrac{1}{{\sqrt x }}$
Or $\sqrt x = 2$
Squaring both sides,
${\left( {\sqrt x } \right)^2} = {\left( 2 \right)^2}$
Which implies x = 4
Also $\dfrac{1}{{\sqrt y }} = v$
Or $\dfrac{1}{3} = \dfrac{1}{{\sqrt y }}$
Or $\sqrt y = 3$
Squaring both sides,
${\left( {\sqrt y } \right)^2} = {\left( 3 \right)^2}$
Which implies y = 9
Therefore, x = 4, y = 9 is the solution of the given equations.

$\left( {iii} \right)\dfrac{4}{x} + 3y = 14;\dfrac{3}{x} - 4y = 23$
Solution- $
  \dfrac{4}{x} + 3y = 14..........(1) \\
  \dfrac{3}{x} - 4y = 23............(2) \\
 $
Let us assume $\dfrac{1}{x} = u$ in equation 1 and equation 2.
So our equation becomes
4u + 3y = 14…………. (3)
3u – 4y = 23………... (4)
Now new formed equations are-
4u + 3y = 14…………. (3)
3u – 4y = 23………... (4)
Using equation 3, we will find the value of u-
4u = 14 – 3y
Or $u = \dfrac{{14 - 3y}}{4}$
Now we will put the value of u in equation 4-
$3\left( {\dfrac{{14 - 3y}}{4}} \right) - 4y = 23$
Multiplying both sides by 4,
We get-
$4 \times 3\left( {\dfrac{{14 - 3y}}{4}} \right) - 4 \times 4y = 4 \times 23$
Or 3(14 – 3y) – 16y = 92
Or 42 – 9y – 16y = 92
Or -25y = 50
Or y = -2
Now we will put the value of y in equation 3-
4u + 3y = 14
Or 4u + 3(-2) = 14
Or 4u – 6 = 14
Or $u = \dfrac{{20}}{4}$
Or u = 5
But $u = \dfrac{1}{x}$
Or $5 = \dfrac{1}{x}$
Or $x = \dfrac{1}{5}$
Hence $x = \dfrac{1}{5}$, y = -2 is the solution of the given equation.

  $\left( {iv} \right)\dfrac{5}{{x - 1}} + \dfrac{1}{{y - 2}} = 2;\dfrac{6}{{x - 1}} - \dfrac{3}{{y - 2}} = 1$
Solution - $
  \dfrac{5}{{x - 1}} + \dfrac{1}{{y - 2}} = 2.........(1) \\
  \dfrac{6}{{x - 1}} - \dfrac{3}{{y - 2}} = 1.........(2) \\
 $
let us assume $\dfrac{1}{{x - 1}} = u$ and $\dfrac{1}{{y - 2}} = v$
put the value of u and v in equations 1 and 2-
5u + v = 2………… (3)
6u – 3v = 1……… (4)
From equation 3, we get –
5u + v = 2
V = 2 – 5u
Putting the value of v in equation 4, we get-
6u – 3(2 – 5u) = 1
Or 6u – 6 + 15u = 1
Or 6u + 15u = 1 + 6
Or 21u = 7
Or $u = \dfrac{7}{{21}}$
Or $u = \dfrac{1}{3}$
Putting the value of u in equation 3, we get-
$5\left( {\dfrac{1}{3}} \right) + v = 2$
Or $\dfrac{5}{3} + v = 2$
Or $v = \dfrac{1}{3}$
Hence we get $u = \dfrac{1}{3}$and $v = \dfrac{1}{3}$
But we need to find the value of x and y,
We know that $u = \dfrac{1}{{x - 1}}$ and $u = \dfrac{1}{3}$
Equating both the values we get-
$\dfrac{1}{3} = \dfrac{1}{{x - 1}}$
Or x – 1 = 3
Or x = 4
Similarly we know that $v = \dfrac{1}{{y - 2}}$ and $v = \dfrac{1}{3}$
Equating both values we get-
$\dfrac{1}{3} = \dfrac{1}{{y - 2}}$
Or y – 2 = 3
Or y = 5
Therefore, x = 4 and y = 5 is the solution of our given equations.

$\left( v \right)\dfrac{{7x - 2y}}{{xy}} = 5;\dfrac{{8x + 7y}}{{xy}} = 15$
Solution- Here we are given that $\dfrac{{7x - 2y}}{{xy}} = 5$
Or $\dfrac{7}{y} - \dfrac{2}{x} = 5$
Or $\dfrac{{ - 2}}{x} + \dfrac{7}{y} = 5........(1)$
And $\dfrac{{8x + 7y}}{{xy}} = 15$
Or $\dfrac{{8x}}{{xy}} + \dfrac{{7y}}{{xy}} = 15$
Or $\dfrac{8}{y} + \dfrac{7}{x} = 15$
Or $\dfrac{7}{x} + \dfrac{8}{y} = 15...........(2)$
Let us assume $\dfrac{1}{x} = u,\dfrac{1}{y} = v$
Our equations become-
-2u + 7v = 5……… (3)
7u + 8v = 15………. (4)
Hence we solve -
-2u + 7v = 5……… (3)
7u + 8v = 15………. (4)
Using equation 3, we get-
7u = 5 + 2u
Or $v = \dfrac{{5 + 2u}}{7}$
Putting the value of v in equation 4-
$7u + 8\left( {\dfrac{{5 + 2u}}{7}} \right) = 15$
Multiplying both sides by 7-
$7 \times 7u + 7 \times 8\left( {\dfrac{{5 + 2u}}{7}} \right) = 7 \times 15$
Or 49u + 40 + 16u= 105
Or 49u + 16u = 105 – 40
Or 65u = 65
Or u = 1
Putting the value of u in equation 3,
-2(1) + 7v = 5
Or -2 + 7v = 5
Or 7v = 7
Or v = 1
Hence u = 1, v = 1
But we need to find x and y-
We know that $u = \dfrac{1}{x}$
Or $1 = \dfrac{1}{x}$
Or x = 1
And $v = \dfrac{1}{y}$
Or $1 = \dfrac{1}{y}$
Or y = 1
Hence x = 1 and y = 1 is the required solution of the given equations.

$\left( {vi} \right)6x + 3y = 6xy;2x + 4y = 5xy$
Solution- Firstly we will divide whole equation by xy -
$\dfrac{{6x + 3y}}{{xy}} = \dfrac{{6xy}}{{xy}}$
Or $\dfrac{6}{y} + \dfrac{3}{x} = 6......(1)$
$\dfrac{{2x + 4y}}{{xy}} = \dfrac{{5xy}}{{xy}}$
Or $\dfrac{2}{y} + \dfrac{4}{x} = 5.........(2)$
So our equations are-
$\dfrac{6}{y} + \dfrac{3}{x} = 6......(1)$
$\dfrac{2}{y} + \dfrac{4}{x} = 5.........(2)$
Let us assume $\dfrac{1}{x} = u,\dfrac{1}{y} = v$
So our equations become-
6v + 3u = 6 ………. (3)
2v + 4u = 5 ……… (4)
Now using equation 3, we get-
$v = \dfrac{{6 - 3u}}{6}$
Putting the value of v in equation 4-
$2\left( {\dfrac{{6 - 3u}}{6}} \right) + 4u = 5$
Multiplying both sides by 3-
$3\left( {\dfrac{{6 - 3u}}{3}} \right) + 3 \times 4u = 3 \times 5$
Or -3u + 12u = 15 – 6
Or 9u = 9
Or u = 1
Putting u = 1 in equation 3-
6v + 3(1) = 6
Or 6v = 3
Or $v = \dfrac{1}{2}$
But we have to find x and y-
$
  u = \dfrac{1}{x} \\
  or{\text{ }}1 = \dfrac{1}{x} \\
  or{\text{ }}x = 1 \\
 $
 $
  v = \dfrac{1}{y} \\
  or{\text{ }}\dfrac{1}{2} = \dfrac{1}{y} \\
  or{\text{ y = 2}} \\
 $
So x = 1 and y = 2 is the required solution of the given equation.

$\left( {vii} \right)\dfrac{{10}}{{x + y}} + \dfrac{2}{{x - y}} = 4;\dfrac{{15}}{{x + y}} - \dfrac{5}{{x - y}} = - 2$
Solution- $\dfrac{{10}}{{x + y}} + \dfrac{2}{{x - y}} = 4.........(1)$
                 $\dfrac{{15}}{{x + y}} - \dfrac{5}{{x - y}} = - 2........(2)$
Let us assume $\dfrac{1}{{x + y}} = u,\dfrac{1}{{x - y}} = v$
So our equations become-
10u + 2v = 4………. (3)
15u – 5v = -2……… (4)
Solving these equations-
Using equation 3,
10u = 4 – 2v
Or $u = \dfrac{{4 - 2v}}{{10}}$
Putting the value of u in equation 4-
$15\left( {\dfrac{{4 - 2v}}{10}} \right) - 5v = - 2$
Multiplying both sides by 2-
$2 \times 3\left( {\dfrac{{4 - 2v}}{2}} \right) - 2 \times 5v = 2 \times - 2$
12 – 6v – 10v = -4
Or -16v = -16
Or v = 1
Putting the value of v in equation 3-
10u + 2(1) = 4
Or 10u + 2 = 4
Or 10u = 2
Or $u = \dfrac{1}{5}$
But we need to find x and y-
$
  u = \dfrac{1}{{x + y}} \\
  or{\text{ }}\dfrac{1}{5} = \dfrac{1}{{x + y}} \\
 $
x + y = 5…… (5)
$
  v = \dfrac{1}{{x - y}} \\
  or{\text{ 1 = }}\dfrac{1}{{x - y}} \\
 $
x – y = 1……. (6)
Adding equation 5 and 6, we get-
(x + y) + (x – y) = 6
2x = 6
Or x = 3
Putting the value of x in equation 5-
3 + y = 5
y = 2
So x = 3 and y = 2 is the solution of our given equations.

$\left( {viii} \right)\dfrac{1}{{3x + y}} + \dfrac{1}{{3x - y}} = \dfrac{3}{4};\dfrac{1}{{2\left( {3x + y} \right)}} - \dfrac{1}{{2\left( {3x - y} \right)}} = \dfrac{{ - 1}}{8}$
Solution- $\dfrac{1}{{3x + y}} + \dfrac{1}{{3x - y}} = \dfrac{3}{4}........(1)$
                 $\dfrac{1}{{2\left( {3x + y} \right)}} - \dfrac{1}{{2\left( {3x - y} \right)}} = \dfrac{{ - 1}}{8}$………. (2)
Let us assume $\dfrac{1}{{3x + y}} = u;\dfrac{1}{{3x - y}} = v$
So our equations become-
$u + v = \dfrac{3}{4}$
Or 4(u + v) = 3
Or 4u + 4v = 3………… (3)
And $\dfrac{1}{2}u - \dfrac{1}{2}v = \dfrac{{ - 1}}{8}$
Or $\dfrac{{u - v}}{2} = \dfrac{{ - 1}}{8}$
Or 4(u – v) = -1
Or 4u – 4v = -1………. (4)
Adding equation 3 and 4, we get
(4u + 4v) + (4u – 4v) = 3 + (-1)
Or 8u = 2
Or $u = \dfrac{1}{4}$
Putting the value of u in equation 3-
$4 \times \dfrac{1}{4} + 4v = 3$
Or 1 + 4v = 3
Or 4v = 2
Or $v = \dfrac{1}{2}$
Hence $u = \dfrac{1}{4},v = \dfrac{1}{2}$
But we need to find x and y-
We know $u = \dfrac{1}{{3x + y}}$
  Or $\dfrac{1}{4} = \dfrac{1}{{3x + y}}$
Or 3x + y = 4………. (5)
$v = \dfrac{1}{{3x - y}}$
Or $\dfrac{1}{2} = \dfrac{1}{{3x - y}}$
Or 3x – y = 2……… (6)
Adding equation5 and equation 6-
(3x + y) + (3x – y) = 6
Or 6x = 6
 Or x = 1
Putting the value of x in equation 5, we get-
3(1) + y = 4
Or 3 + y = 4
Or y = 4 – 3
Or y = 1
So x = 1 and y = 1 is the solution of the given equation.

Note- Here in order to solve these type of questions, we used a substitution method but we can use other methods also like elimination method and augmented matrices. Also one must know that there is a difference between the methods of solving linear equations in one variable and linear equations in two variables.