SELINA Solution Class 9 Simultaneous (Linear) Equation (Including problems) Chapter 6 Exercise 6D

Question 1

Solve :
${\displaystyle \frac{9}{x}-\frac{4}{y}=8}$

${\displaystyle \frac{13}{x}+\frac{7}{y}=101}$

Sol:

${\displaystyle \frac{9}{x}-\frac{4}{y}=8}$                             ....(1)

${\displaystyle \frac{13}{x}+\frac{7}{y}=101}$                      .....(2)

Multiplying equation no. (1) by 7 and (2) by 4.
${\displaystyle \frac{63}{x}-\frac{28}{y}=56}$                       .....(3)

${\displaystyle \frac{52}{x}+\frac{28}{y}=404}$                    .....(4)

Adding equation (3) and (4)
        ${\displaystyle \frac{63}{x}-\frac{28}{y}=56}$  

 +    ${\displaystyle \frac{52}{x}+\frac{28}{y}=404}$ 
                                          
      ${\displaystyle \frac{115}{x}}$ = 460
      ${\displaystyle x=\frac{115}{460}=x=\frac{1}{4}}$

From (1)
9 x ${\displaystyle \left(\frac{4}{1}\right)-\frac{4}{y}=8}$
- ${\displaystyle \frac{4}{y}=-28}$
${\displaystyle y=\frac{1}{7}}$

Question 2

Solve the pairs of equations :
${\displaystyle \frac{3}{x}+\frac{2}{y}=10}$

${\displaystyle \frac{9}{x}-\frac{7}{y}=10.5}$

Sol:

${\displaystyle \frac{3}{x}+\frac{2}{y}=10}$                  .....(1)

${\displaystyle \frac{9}{x}-\frac{7}{y}=10.5}$                .....(2)

Multiplying equation (1) by 3, We get
${\displaystyle \frac{9}{x}+\frac{6}{y}=30}$                 ......(3)

Subtracting (2) from (3), We get
       ${\displaystyle \frac{9}{x}+\frac{6}{y}=30}$ 
-     ${\displaystyle \frac{9}{x}-\frac{7}{y}=10.5}$
       -      +        -         
               ${\displaystyle \frac{13}{y}=19.5}$

                ${\displaystyle y=\frac{13}{19.5}=\frac{2}{3}}$

From (1),
${\displaystyle \frac{3}{x}+\frac{2\times 3}{2}=10}$
⇒ ${\displaystyle \frac{3}{x}+3=10}$
⇒ ${\displaystyle \frac{3}{x}=7}$
⇒ x = ${\displaystyle \frac{3}{7}}$

Question 3

Solve :
5x + ${\displaystyle \frac{8}{y}}$ = 19

3x - ${\displaystyle \frac{4}{y}}$ = 7

Sol:

5x + ${\displaystyle \frac{8}{y}}$ = 19                         ....(1)

3x - ${\displaystyle \frac{4}{y}}$ = 7                            .....(2)
Multiplying equation (2) by 2, We get,
6x - ${\displaystyle \frac{8}{y}}$ = 14                         ......(3)

Adding (1) and (3), We get,
    5x + ${\displaystyle \frac{8}{y}}$ = 19

+  6x - ${\displaystyle \frac{8}{y}}$ = 14
                              
    11x = 33
        x = 3

Substituting x = 3 in equation (1), We get
5(3) + ${\displaystyle \frac{8}{y}}$ = 19
15 + ${\displaystyle \frac{8}{y}}$ = 19
${\displaystyle \frac{8}{y}}$ = 19 - 15
y = ${\displaystyle \frac{8}{4}}$ 
y = 2

Question 4

Solve :
4x + ${\displaystyle \frac{6}{y}}$ = 15 and 3x - ${\displaystyle \frac{4}{y}}$ = 7. Hence, find a if y = ax - 2.

Solve :
4x + ${\displaystyle \frac{6}{y}}$ = 15 and 3x - ${\displaystyle \frac{4}{y}}$ = 7. Hence, find a if y = ax - 2.

Sol:

4x + ${\displaystyle \frac{6}{y}}$ = 15                           .....(1)

3x - ${\displaystyle \frac{4}{y}}$ = 7                              .....(2)
Multiplying (1) by 4 and (2) by 6
16x + ${\displaystyle \frac{24}{y}}$ = 60                       ....(3)

18x - ${\displaystyle \frac{24}{y}}$ = 42                        ....(4)
Adding (3) and (4), We get
        16x + ${\displaystyle \frac{24}{y}}$ = 60

+      18x - ${\displaystyle \frac{24}{y}}$ = 42
                                         
         34x = 102
            x = 3
Substituting x = 3 in (1), We get
4(3) + ${\displaystyle \frac{6}{y}}$ = 15

⇒ ${\displaystyle \frac{6}{y}}$ = 15 - 12

⇒ y = ${\displaystyle \frac{6}{3}}$ = 2
Now,
y = ax - 2
2 = a(3) - 2
3a = 4 
a = ${\displaystyle \frac{4}{3} =1\frac{1}{3}}$

Question 5

Solve :
${\displaystyle \frac{3}{x}-\frac{2}{y}=0\text{and}\frac{2}{x}+\frac{5}{y}=19}$ Hence, find 'a' if y = ax + 3.

Sol:

${\displaystyle \frac{3}{x}-\frac{2}{y}=0}$                       ......(1)

${\displaystyle \frac{2}{x}+\frac{5}{y}=19}$                   .......(2)

Multiplying equation no. (1) by 5 and (2) by 2.
${\displaystyle \frac{15}{x}-\frac{10}{y}=0}$                  ........(3)

${\displaystyle \frac{4}{x}+\frac{10}{y}=38}$                 ........(4)

Adding (3) and (4),
        ${\displaystyle \frac{15}{x}-\frac{10}{y}=0}$                  

  +    ${\displaystyle \frac{4}{x}+\frac{10}{y}=38}$ 
                                         
 ${\displaystyle \frac{19}{x}=38}$
  x = ${\displaystyle \frac{1}{2}}$
From (1)
${\displaystyle 3\left(\frac{1}{2}\right)-\frac{2}{y}=0}$
y = ${\displaystyle \frac{1}{3}}$

∴ y = ax + 3
${\displaystyle \frac{1}{3}=a\left(\frac{1}{2}\right)+3}$
${\displaystyle \frac{a}{2}=-\frac{8}{3}}$
${\displaystyle a=-\frac{16}{3}}$

Question 6.1

Solve :
${\displaystyle \frac{20}{x+y}+\frac{3}{x-y}=7}$

${\displaystyle \frac{8}{x-y}-\frac{15}{x+y}=5}$

Sol:

${\displaystyle \frac{20}{x+y}+\frac{3}{x-y}=7}$          .....(1)

${\displaystyle \frac{8}{x-y}-\frac{15}{x+y}=5}$          ......(2)

Multiplying equation no (1) by 8 and (2) by 3.
${\displaystyle \frac{160}{x+y}+\frac{24}{x-y}=56}$    ......(3)

${\displaystyle -\frac{45}{x+y}+\frac{24}{x-y}=15}$     ......(4)

Subtracting equation (4) from (3)
      ${\displaystyle \frac{160}{x+y}+\frac{24}{x-y}=56}$

-   ${\displaystyle -\frac{45}{x+y}+\frac{24}{x-y}=15}$
                                                          
      ${\displaystyle \frac{205}{x+y}=41}$

x + y = 5                                    .......(5)

From (1)
${\displaystyle \frac{20}{5}+\frac{3}{x-y}=7}$

${\displaystyle \frac{3}{x-y}=3}$
x - y = 1                                   ......(6)

Adding equation (5) and (6)
      x + y = 5
+    x - y = 1  
         2x = 6
           x = 3

From (5)
3 + y = 5
y = 5 - 3
y = 2

Question 6.2

Solve : 
${\displaystyle \frac{34}{3x+4y}+\frac{15}{3x-2y}=5}$

${\displaystyle \frac{25}{3x-2y}-\frac{8.50}{3x+4y}=4.5}$

Sol:

Let a = 3x + 4y and b = 3x - 2y
∴ ${\displaystyle \frac{34}{3x+4y}+\frac{15}{3x-2y}=5}$

⇒ ${\displaystyle \frac{34}{a}+\frac{15}{b}=5}$                 .....(1)

${\displaystyle \frac{25}{3x-2y}-\frac{8.50}{3x+4y}=4.5}$

⇒ ${\displaystyle -\frac{8.50}{a}+\frac{25}{b}=4.5}$        .....(2)
Multiply equation (2) by 4, We get :
${\displaystyle -\frac{34}{a}+\frac{100}{b}=18}$             ......(3)

Adding equation (1) and (3)
        ${\displaystyle -\frac{34}{a}+\frac{100}{b}=18}$ 

+      ${\displaystyle \frac{34}{a}+\frac{15}{b}=5}$
                                              
          ${\displaystyle \frac{115}{b}=23}$
                  b = 5
3x - 2y = 5                             .......(4)
Substituting b = 5 in equation (1), We get
${\displaystyle \frac{34}{a}+\frac{15}{b}=5}$ 

${\displaystyle \frac{34}{a}+\frac{15}{5}=5}$ 

${\displaystyle \frac{34}{a}=2}$
2a = 34
a = 17

3x + 4y = 17                            ......(5)
Subtracting equation (5) from (4), We get :
        3x - 2y = 5
-      3x + 4y = 17  
      -     -         -     
            - 6y = - 12
                y = 2
Substituting y = 2 in equation (4), We get
3x - 2(2) = 5
3x = 9
x = 3
∴ Solution is x = 3 and y = 2.

Question 7.1

Solve :
x + y = 2xy
x - y = 6xy

Sol:

x + y = 2xy                       .....(1)
x - y = 6xy                        ......(2)
Adding equation (1) and (2)
     x + y = 2xy
+   x - y = 6xy 
      2x = 8xy
      2 = 8y
      y = ${\displaystyle \frac{1}{4}}$
From (1)
${\displaystyle x+\frac{1}{4}=2\times \left(\frac{1}{4}\right)}$
${\displaystyle \frac{1}{2}x=-\frac{1}{4}}$
${\displaystyle x=-\frac{1}{2}}$

Question 7.2

Solve :
x+ y = 7xy
2x - 3y = - xy

Sol:

x + y = 7xy                           ...(1)
2x - 3 = - xy                          ...(2)
Multiplying equation no. (1) by 3.
3x + 3y = 21xy                     ....(3)
Adding equation (3) and (2)
     3x + 3y = 21xy
+   2x - 3y  = - xy  
      5x = 20xy
        y = ${\displaystyle \frac{1}{4}}$
From (1)
x + ${\displaystyle \frac{1}{4}=7x\left(\frac{1}{4}\right)}$

${\displaystyle \frac{1}{4}=\frac{3}{4}x}$

x = ${\displaystyle \frac{1}{3}}$

Question 8

Solve :
${\displaystyle \frac{a}{x}-\frac{b}{y}=0}$

${\displaystyle \frac{a{b}^2}{x}+\frac{a^{2}b}{y}=a^2+b^2}$

Sol:

Given equation are ${\displaystyle \frac{a}{x}-\frac{b}{y}=0\text{and}\frac{a{b}^2}{x}+\frac{a^{2}b}{y}=a^2+b^2}$

Taking ${\displaystyle \frac{1}{x}=u\text{and}\frac{1}{y}=v}$, the above system of equations become
au - bv + 0 = 0
ab²u + a²bv - ( a² + b² ) = 0
By cross-multiplication, we have
${\displaystyle \frac{u}{-b\times [-(a^2+b^2)]-a^{2}b\times 0}=\frac{-v}{a\times [-(a^2+b^2)]-a{b}^2\times 0}=\frac{1}{a\times a^{2}b-a{b}^2\times (-b)}}$

⇒ ${\displaystyle \frac{u}{b(a^2+b^2)}=\frac{-v}{-a(a^2+b^2)}=\frac{1}{a^{3}b+a{b}^3}}$

⇒ ${\displaystyle \frac{u}{b(a^2+b^2)}=\frac{v}{a(a^2+b^2)}=\frac{1}{ab(a^2+b^2)}}$

⇒ ${\displaystyle u=\frac{b(a^2+b^2)}{ab(a^2+b^2)} \text{and} v=\frac{a(a^2+b^2)}{ab(a^2+b^2)}}$

⇒ u = ${\displaystyle \frac{1}{a}}$  and v = ${\displaystyle \frac{1}{b}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{1}{a} \text{and}\frac{1}{y}=\frac{1}{b}}$ 

⇒ x =a and y = b

Question 9

Solve : 
${\displaystyle \frac{2xy}{x+y}=\frac{3}{2}}$

${\displaystyle \frac{xy}{2x-y}=-\frac{3}{10}}$
x + y ≠ 0 and 2x - y ≠ 0 

Sol:

${\displaystyle \frac{2xy}{x+y}=\frac{3}{2}}$

⇒ ${\displaystyle \frac{x+y}{xy}=\frac{4}{3}}$

⇒ ${\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{4}{3}}$              .....(1)

${\displaystyle \frac{xy}{2x-y}=-\frac{3}{10}}$

⇒ ${\displaystyle \frac{2x-y}{xy}=-\frac{10}{3}}$

⇒ ${\displaystyle -\frac{1}{x}+\frac{2}{y}=-\frac{10}{3}}$        ......(2)

Let ${\displaystyle \frac{1}{x}=u\text{and}\frac{1}{y}=v}$
Then, equations (1) and (2) become
u + v = ${\displaystyle \frac{4}{3} \text{and} -u+2v=-\frac{10}{3}}$

⇒ 3u + 3v = 4 and -3u + 6v = -10
Adding, We have
9v = - 6
⇒ v = ${\displaystyle -\frac{6}{9}=-\frac{2}{3}}$

⇒ ${\displaystyle \frac{1}{y}=-\frac{2}{3}}$
⇒ y = ${\displaystyle -\frac{3}{2}}$

Substituting y = ${\displaystyle -\frac{3}{2}}$ in (1), We have
${\displaystyle \frac{1}{x}-\frac{2}{3}=\frac{4}{3}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{6}{3}=2}$

⇒ x = ${\displaystyle \frac{1}{2}}$
Hence, x = ${\displaystyle \frac{1}{2} \text{and}y=-\frac{3}{2}}$

Question 10

Solve :
${\displaystyle \frac{3}{2x}+\frac{2}{3y}=-\frac{1}{3}}$

${\displaystyle \frac{3}{4x}+\frac{1}{2y}=-\frac{1}{8}}$

Sol:

Given equations are ${\displaystyle \frac{3}{2x}+\frac{2}{3y}=-\frac{1}{3}\text{and}\frac{3}{4x}+\frac{1}{2y}=-\frac{1}{8}}$

Let ${\displaystyle \frac{1}{x}=u\text{and}\frac{1}{y}=v}$
Then, the system of equations become
${\displaystyle \frac{3}{2}u+\frac{2}{3}v=-\frac{1}{3} \text{and}\frac{3}{4}u+\frac{1}{2}v=-\frac{1}{8}}$

⇒ ${\displaystyle \frac{9u+4v}{6}=-\frac{1}{3}\text{and} \frac{3u+2v}{4}=-\frac{1}{8}}$

⇒ 27u + 12v = -6 and 24u + 16v = -4
⇒ 27u + 12v + 6 = 0 and 24u + 16v + 4 = 0

⇒ ${\displaystyle \frac{u}{12\times 4-16\times 6}=\frac{-v}{27\times 4-24\times 6}=\frac{1}{27\times 16-24\times 12}}$

⇒ ${\displaystyle \frac{u}{48-96}=\frac{-v}{108-144}=\frac{1}{432-288}}$

⇒ ${\displaystyle \frac{u}{-48}=\frac{-v}{-36}=\frac{1}{144}}$

⇒ ${\displaystyle \frac{u}{-48}=\frac{v}{36}=\frac{1}{144}}$

⇒ ${\displaystyle u=\frac{-48}{144}=\frac{1}{3} \text{and}v=\frac{36}{144}=\frac{1}{4}}$

⇒ ${\displaystyle \frac{1}{x}=-\frac{1}{3}\text{and}\frac{1}{y}=\frac{1}{4}}$

⇒ x = - 3 and y = 4.