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

Question 1

Solve, using cross-multiplication :
4x + 3y = 17
3x - 4y + 6 = 0

Sol:

Given equation are 4x + 3y = 17 and 3x - 4y + 6 = 0
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = - 4, c2 = 6

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{3\times 6-(-4)\times (-17)}{4\times (-4)-3\times 3} \text{and} y=\frac{-17\times 3-6\times 4}{4\times (-4)-3\times 3}}$

⇒ x = ${\displaystyle \frac{18-68}{-16-9} \text{and}y=\frac{-51-24}{-16-9}}$

⇒ ${\displaystyle x=\frac{-50}{-25}\text{and}y=\frac{-75}{-25}}$

⇒ x = 2 and y = 3.

Question 2

Solve, using cross-multiplication :
3x + 4y = 11
2x + 3y = 8

Sol:

Given equations are 3x + 4y = 11 and 2x + 3y = 8
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we have
a₁ = 3, b₁ = 4, c₁ = -11 and a₂ = 2, b₂ = 3, c₂ = - 8

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{4\times (-8)-3\times (-11)}{3\times 3-2\times 4} \text{and} y=\frac{-11\times 2-(-8)\times 3}{3\times 3-2\times 4}}$

⇒ ${\displaystyle x=\frac{-32+33}{9-8}\text{and}y=\frac{-22+24}{9-8}}$

⇒ x = 1 and y = 2

Question 3

Solve, using cross-multiplication :
6x + 7y - 11 = 0
5x + 2y = 13

Sol:

Given equation are 6x + 7y - 11 = 0 and 5x + 2y = 13
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 6, b₁ = 7, c₁ = -11 and a₂ = 5, b₂ = 2, c₂ = -13

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{7\times (-13)-2\times (-11)}{6\times 2-5\times 7} \text{and} y=\frac{-11\times 5-(-13)\times 6}{6\times 2-5\times 7}}$

⇒ x = ${\displaystyle \frac{-91+22}{12-35} \text{and}y=\frac{-55+78}{12-35}}$

⇒ x = ${\displaystyle \frac{-69}{-23} \text{and}y=\frac{23}{-23}}$
⇒ x = 3 and y = -1

Question 4

Solve, using cross-multiplication :
5x + 4y + 14 = 0
3x = -10 - 4y

Sol:

Given equation are 5x + 4y + 14 = 0 and 3x = -10 - 4y
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have
a₁ = 5, b₁ = 4, c₁ = 14 and a₂ = 3, b₂ =  4, c₂ = 10

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{4\times 10-4\times 14}{5\times 4-3\times 4} \text{and} y=\frac{14\times 3-10\times 5}{5\times 4-3\times 4}}$

⇒ x = ${\displaystyle \frac{40-56}{20-12}\text{and}y=\frac{42-50}{20-12}}$

⇒ x = ${\displaystyle -\frac{16}{8}\text{and}y=-\frac{8}{8}}$

⇒ x = -2 and y = -1

Question 5

Solve, using cross-multiplication :
x - y + 2 = 0
7x + 9y = 130

Sol:

Given equation are x - y + 2 = 0 and 7x + 9y = 130
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 1, b₁ = -1, c₁ = 2 and a₂ = 7, b₂ = 9, c₂ = -130

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{-1\times (-130)-9\times 2}{1\times 9-7\times (-1)} \text{and} y=\frac{2\times 7-(-130)\times 1}{1\times 9-7\times (-1)}}$

⇒ x = ${\displaystyle \frac{130-18}{9+7} \text{and} y=\frac{14+130}{9+7}}$

⇒ x = ${\displaystyle \frac{112}{16} \text{and}y=\frac{144}{16}}$

⇒ x = 7 and y = 9.

Question 6

Solve, using cross-multiplication :
4x - y = 5
5y - 4x = 7

Sol:

Given equation are 4x - y = 5 and 5y - 4x = 7
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 41, b₁ = - 1, c₁ = - 5 and a₂ = - 4, b₂ = 5, c₂ = - 7

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{-1\times (-7)-5\times (-5)}{4\times 5-(-4)\times (-1)} \text{and} y=\frac{(-5)\times (-4)-(-7)\times 4}{4\times 5-(-4)\times (-1)}}$

⇒ ${\displaystyle x=\frac{7+25}{20-4} \text{and}y=\frac{20+28}{20-4 }}$

⇒ x = ${\displaystyle \frac{32}{16} \text{and}y=\frac{48}{16}}$

⇒ x = 2 and y = 3

Quesiton 7

Solve, using cross-multiplication :
4x - 3y = 0
2x + 3y = 18

Sol:

Given equation are 4x - 3y = 0 and 2x + 3y = 18
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 4, b₁ = - 3, c₁ = 0 and a₂ = 2, b₂ = 3, c₂ = -18

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{-3\times (-18)-3\times 0}{4\times 3-2\times (-3)} \text{and} y=\frac{0\times 2-(-18)\times 4}{4\times 3-2\times (-3)}}$

⇒ ${\displaystyle x=\frac{54-0}{12+6} \text{and}y=\frac{0+72}{12+6}}$

⇒ x = ${\displaystyle \frac{54}{18}\text{and}y=\frac{72}{18}}$

⇒ x = 3 and y = 4.

Question 8

Solve, using cross-multiplication :
8x + 5y = 9
3x + 2y = 4

Sol:

Given equation are 8x + 5y = 9 and 3x + 2y = 4
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 8, b₁ = 5, c₁ = -9 and a₂ = 3, b₂ = 2, c₂ = - 4

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{5\times (-4)-2\times (-9)}{8\times 2-3\times 5} \text{and} y=\frac{-9\times 3-(-4)\times 8}{8\times 2-3\times 5}}$

⇒ x = ${\displaystyle \frac{-20+18}{16-15}\text{and}y=\frac{-27+32}{16-15}}$

⇒ x = ${\displaystyle -\frac{2}{1}\text{and}y=\frac{5}{1}}$
⇒ x = - 2 and y = 5.

Question 9

Solve, using cross-multiplication :
4x - 3y - 11 = 0
6x + 7y - 5 = 0

Sol:

Given equation are 4x - 3y - 11 = 0 and 6x + 7y - 5 = 0
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 4, b₁ = - 3, c₁ = -11 and a₂ = 6, b₂ = 7, c₂ = - 5

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{-3\times (-5)-7\times (-11)}{4\times 7-6\times (-3)} \text{and} y=\frac{-11\times 6-(-5)\times 4}{4\times 7-6\times (-3)}}$

⇒ x = ${\displaystyle \frac{15+77}{28+18}\text{and}y=\frac{-66+20}{28+18}}$

⇒ x = ${\displaystyle \frac{92}{46}\text{and}y=-\frac{46}{46}}$

⇒ x = 2 and y = - 1

Question 10

Solve, using cross-multiplication :
4x + 6y = 15
3x - 4y = 7

Sol:

Given equation are 4x + 6y = 15 and 3x - 4y = 7
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 4, b₁ = 6, c₁ = -15 and a₂ = 3, b₂ = - 4, c₂ = -7

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{6\times (-7)-(-4)\times (-15)}{4\times (-4)-3\times 6} \text{and} y=\frac{-15\times 3-(-7)\times 4}{4\times (-4)-3\times 6}}$

⇒ x = ${\displaystyle \frac{-42-60}{-16-18} \text{and}y=\frac{-45+28}{-16-18}}$

⇒ x ${\displaystyle \frac{-102}{-34} \text{and}y=\frac{-17}{-34}}$

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

Question 11

Solve, using cross-multiplication :
0.4x - 1.5y = 6.5
0.3x + 0.2y = 0.9

Sol:

Given equation are 0.4x - 1.5y = 6.5 and 0.3x + 0.2y = 0.9
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 0.4, b₁ = -1.5, c₁ = -6.5 and a₂ = 0.3, b₂ = 0.2, c₂ = 0.9

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{(-1.5)\times (-0.9)-\left(0.2\right)\times (-6.5)}{0.4\times \left(0.2\right)-\left(0.3\right)\times (-1.5)} \text{and} y=\frac{(-6.5)\times \left(0.3\right)-(-0.9)\times \left(0.4\right)}{0.4\times \left(0.2\right)-\left(0.3\right)\times (-1.5)}}$

⇒ x = ${\displaystyle \frac{1.35+1.3}{0.08+0.45} \text{and}y=\frac{-1.95+0.36}{0.08+0.45}}$

⇒ x = ${\displaystyle \frac{2.65}{0.53}\text{and}y=\frac{-1.59}{0.53}}$

⇒ x = 5 and y = - 3

Question 12

Solve, using cross-multiplication :
√2x - √3y = 0
√5x + √2y = 0

Sol;

Given equation are √2x - √3y = 0 and √5x + √2y = 0
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = √2, b₁ = √3, c₁ = 0 and a₂ = √5, b₂ = √2, c₂ = 0

Now, x = ${\displaystyle \frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1} \text{and} y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}}$ 

⇒ x = ${\displaystyle \frac{(-\sqrt{3})\times 0-\sqrt{2}\times 0}{\sqrt{2}\times \sqrt{2}-\sqrt{5}\times (-\sqrt{3})} \text{and} y=\frac{0\times \sqrt{5}-0\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}-\sqrt{5}\times (-\sqrt{3})}}$

⇒ x = ${\displaystyle \frac{0}{2+\sqrt{15}}\text{and}y=\frac{0}{2+\sqrt{15}}}$

⇒ x = 0 and y = 0.