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Q1.

Which one of the following options is true, and why?
$y=3x+5$ has (i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions.

Answer 1. (iii) infinitely many solutions.
Reason: $y=3x+5$ is a linear equation in two variables $x$ and $y$. For every real number $x$ you choose there is a corresponding $y$ given by $3x+5$. So there are infinitely many ordered-pair solutions $(x,y)$.

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Q2.

Write four solutions for each of the following equations.

(i) $2x+y=7$
Four solutions (pick convenient $x$-values and solve for $y$):

• $x=0\Rightarrow y=7$ → $(0,7)$

• $x=1\Rightarrow y=5$ → $(1,5)$

• $x=2\Rightarrow y=3$ → $(2,3)$

• $x=3\Rightarrow y=1$ → $(3,1)$

(ii) $\pi x+y=9$
Four solutions:

• $x=0\Rightarrow y=9$ → $(0,9)$

• $x=1\Rightarrow y=9-\pi$ → $(1,9-\pi )$

• $x=2\Rightarrow y=9-2\pi$ → $(2,9-2\pi )$

• $x=3\Rightarrow y=9-3\pi$ → $(3,9-3\pi )$

(Any real $x$ gives a corresponding real $y$.)

(iii) $x=4y$
Four solutions (pick $y$-values and compute $x$):

• $y=0\Rightarrow x=0$ → $(0,0)$

• $y=1\Rightarrow x=4$ → $(4,1)$

• $y=2\Rightarrow x=8$ → $(8,2)$

• $y=-1\Rightarrow x=-4$ → $(-4,-1)$

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Q3.

Check which of the following are solutions of $x-2y=4$ and which are not:
$(i)(0,2)(ii)(2,0)(iii)(4,0)(iv)(2,2)(v)(1,1)$

Check by substitution $x-2y$:

• (i) $(0,2)$: $0-2\cdot 2=0-4=-4\mathrm{\ne }4$→ Not a solution.

• (ii) $(2,0)$: $2-2\cdot 0=2\mathrm{\ne }4$ → Not a solution.

• (iii) $(4,0)$: $4-2\cdot 0=4$ → Solution.

• (iv) $(2,2)$: $2-2\cdot 2=2-4=-2\mathrm{\ne }4$ → Not a solution.

• (v) $(1,1)$: $1-2\cdot 1=1-2=-1\mathrm{\ne }4$ → Not a solution.

So only $(4,0)$ is a solution of $x-2y=4$

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Q4.

Find the value of $k$, if $x=2,y=1$ is a solution of $2x+3y=k$

Answer 4. Substitute $x=2,y=1$: $k=2(2)+3(1)=4+3=7$