Class 9-NCERT Solutions-Chapter-04-Linear Equations in two variables

Solution:

Let the cost of a notebook and a pen be \( x \) and \( y \) respectively.

Cost of notebook = 2 × Cost of pen

\[ x = 2y \]

\[ x - 2y = 0 \]

Solution:

(i) \( 2x+3y=9.3\overline{5} \)

Rewriting as \( 2x+3y-9.3\overline{5}=0 \)

Comparing with \( ax+by+c=0 \): \( a=2, b=3, c=-9.3\overline{5} \)

(ii) \( x-\frac{y}{5}-10=0 \)

Comparing with \( ax+by+c=0 \): \( a=1, b=-\frac{1}{5}, c=-10 \)

(iii) \( -2x+3y=6 \)

Rewriting as \( -2x+3y-6=0 \)

Comparing with \( ax+by+c=0 \): \( a=-2, b=3, c=-6 \)

(iv) \( x=3y \)

Rewriting as \( 1x-3y+0=0 \)

Comparing with \( ax+by+c=0 \): \( a=1, b=-3, c=0 \)

(v) \( 2x=-5y \)

Rewriting as \( 2x+5y+0=0 \)

Comparing with \( ax+by+c=0 \): \( a=2, b=5, c=0 \)

(vi) \( 3x+2=0 \)

Rewriting as \( 3x+0y+2=0 \)

Comparing with \( ax+by+c=0 \): \( a=3, b=0, c=2 \)

(vii) \( y-2=0 \)

Rewriting as \( 0x+1y-2=0 \)

Comparing with \( ax+by+c=0 \): \( a=0, b=1, c=-2 \)

(viii) \( 5=2x \)

Rewriting as \( -2x+0y+5=0 \)

Comparing with \( ax+by+c=0 \): \( a=-2, b=0, c=5 \)

Answer: (iii) infinitely many solutions

Reason: \( y=3x+5 \) is a linear equation in two variables and it has infinite possible solutions. As for every value of \( x \), there will be a value of \( y \) satisfying the above equation and vice-versa.

(i) \( 2x+y=7 \)

• For \( x=0 \): \( 2(0)+y=7 \Rightarrow y=7 \). Solution: (0, 7)
• For \( x=1 \): \( 2(1)+y=7 \Rightarrow y=5 \). Solution: (1, 5)
• For \( x=-1 \): \( 2(-1)+y=7 \Rightarrow y=9 \). Solution: (-1, 9)
• For \( x=2 \): \( 2(2)+y=7 \Rightarrow y=3 \). Solution: (2, 3)

(ii) \( \pi x+y=9 \)

• For \( x=0 \): \( \pi(0)+y=9 \Rightarrow y=9 \). Solution: (0, 9)
• For \( x=1 \): \( \pi(1)+y=9 \Rightarrow y=9-\pi \). Solution: (1, 9-\(\pi\))
• For \( x=2 \): \( \pi(2)+y=9 \Rightarrow y=9-2\pi \). Solution: (2, 9-2\(\pi\))
• For \( x=-1 \): \( \pi(-1)+y=9 \Rightarrow y=9+\pi \). Solution: (-1, 9+\(\pi\))

(iii) \( x=4y \)

• For \( x=0 \): \( 0=4y \Rightarrow y=0 \). Solution: (0, 0)
• For \( y=1 \): \( x=4(1)=4 \). Solution: (4, 1)
• For \( y=-1 \): \( x=4(-1)=-4 \). Solution: (-4, -1)
• For \( x=2 \): \( 2=4y \Rightarrow y=\frac{2}{4}=\frac{1}{2} \). Solution: (2, \(\frac{1}{2}\))

Solution:

(i) (0, 2): Putting \( x=0 \) and \( y=2 \) in the L.H.S: \( 0-2(2)=-4 \neq 4 \). Not a solution.

(ii) (2, 0): Putting \( x=2 \) and \( y=0 \) in the L.H.S: \( 2-2(0)=2 \neq 4 \). Not a solution.

(iii) (4, 0): Putting \( x=4 \) and \( y=0 \) in the L.H.S: \( 4-2(0)=4 \). It is a solution.

(iv) \( (\sqrt{2}, 4\sqrt{2}) \): Putting \( x=\sqrt{2} \) and \( y=4\sqrt{2} \) in the L.H.S: \( \sqrt{2}-2(4\sqrt{2}) = -7\sqrt{2} \neq 4 \). Not a solution.

(v) (1, 1): Putting \( x=1 \) and \( y=1 \) in the L.H.S: \( 1-2(1)=-1 \neq 4 \). Not a solution.

Solution:

Putting \( x=2 \) and \( y=1 \) in the given equation \( 2x+3y=k \):

\[ 2(2)+3(1)=k \Rightarrow 4+3=k \Rightarrow k=7 \]

Therefore, the value of k is 7.