EXERCISE 3.3

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1. Solve the following pair of linear equations by the elimination method and the substitution method.

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(i) $x+y=5$ and $2x-3y=4$

Solution:

From $$ x+y=5 $$ $$ x=5-y $$ Substitute into $$ 2x-3y=4 $$ $$ 2(5-y)-3y=4 $$ $$ 10-2y-3y=4 $$ $$ 10-5y=4 $$ $$ -5y=-6 $$ $$ y=\frac65 $$ Now, $$ x=5-\frac65 $$ $$ x=\frac{19}{5} $$

Verification by Elimination:

Multiply the first equation by $2$. $$ 2x+2y=10 $$ Subtract $$ 2x-3y=4 $$ $$ 5y=6 $$ $$ y=\frac65,\qquad x=\frac{19}{5} $$

Answer:

$$ \boxed{x=\frac{19}{5},\qquad y=\frac65} $$

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(ii) $3x+4y=10$ and $2x-2y=2$

Solution:

From $$ 2x-2y=2 $$ $$ x-y=1 $$ $$ x=y+1 $$ Substitute into $$ 3x+4y=10 $$ $$ 3(y+1)+4y=10 $$ $$ 7y+3=10 $$ $$ 7y=7 $$ $$ y=1 $$ Now, $$ x=1+1=2 $$

Verification by Elimination:

Multiply $$ 2x-2y=2 $$ by $2$. $$ 4x-4y=4 $$ Multiply $$ 3x+4y=10 $$ by $1$. Add, $$ 7x=14 $$ $$ x=2,\qquad y=1 $$

Answer:

$$ \boxed{x=2,\qquad y=1} $$

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(iii) $3x-5y-4=0$ and $9x=2y+7$

Solution:

Write in standard form. $$ 3x-5y=4 $$ $$ 9x-2y=7 $$ From $$ 3x-5y=4 $$ $$ 3x=4+5y $$ $$ x=\frac{4+5y}{3} $$ Substitute into $$ 9x-2y=7 $$ $$ 9\left(\frac{4+5y}{3}\right)-2y=7 $$ $$ 3(4+5y)-2y=7 $$ $$ 12+15y-2y=7 $$ $$ 13y=-5 $$ $$ y=-\frac5{13} $$ Now, $$ x=\frac{4+5\left(-\frac5{13}\right)}3 $$ $$ x=\frac{\frac{52}{13}-\frac{25}{13}}3 $$ $$ x=\frac{27}{39} $$ $$ x=\frac9{13} $$

Verification by Elimination:

Multiply $$ 3x-5y=4 $$ by $3$. $$ 9x-15y=12 $$ Subtract $$ 9x-2y=7 $$ $$ -13y=5 $$ $$ y=-\frac5{13},\qquad x=\frac9{13} $$

Answer:

$$ \boxed{x=\frac9{13},\qquad y=-\frac5{13}} $$

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(iv) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$ and $x-\dfrac{y}{3}=3$

Solution:

Multiply the first equation by $6$. $$ 3x+4y=-6 $$ Multiply the second equation by $3$. $$ 3x-y=9 $$ From $$ 3x-y=9 $$ $$ y=3x-9 $$ Substitute into $$ 3x+4y=-6 $$ $$ 3x+4(3x-9)=-6 $$ $$ 15x-36=-6 $$ $$ 15x=30 $$ $$ x=2 $$ Now, $$ y=3(2)-9 $$ $$ y=-3 $$

Verification by Elimination:

Subtract $$ 3x-y=9 $$ from $$ 3x+4y=-6 $$ $$ 5y=-15 $$ $$ y=-3,\qquad x=2 $$

Answer:

$$ \boxed{x=2,\qquad y=-3} $$

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2. Form the pair of linear equations for the following problems and find their solutions by the elimination method.

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(i) If we add $1$ to the numerator and subtract $1$ from the denominator, a fraction reduces to $1$. It becomes $\dfrac12$ if only $1$ is added to the denominator. What is the fraction?

Solution:

Let the numerator be $x$ and denominator be $y$. Given, $$ \frac{x+1}{y-1}=1 $$ $$ x-y=-2 $$ Also, $$ \frac{x}{y+1}=\frac12 $$ $$ 2x=y+1 $$ $$ 2x-y=1 $$ Subtract the first equation from the second. $$ x=3 $$ Substitute into $$ x-y=-2 $$ $$ 3-y=-2 $$ $$ y=5 $$

Answer:

$$ \boxed{\frac35} $$

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(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution:

Let Nuri's present age be $x$ years and Sonu's present age be $y$ years. Five years ago, $$ x-5=3(y-5) $$ $$ x-3y=-10 $$ Ten years later, $$ x+10=2(y+10) $$ $$ x-2y=10 $$ Subtract the equations. $$ y=20 $$ Substitute into $$ x-2y=10 $$ $$ x-40=10 $$ $$ x=50 $$

Answer:

$$ \boxed{\text{Nuri}=50\text{ years},\qquad \text{Sonu}=20\text{ years}} $$

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(iii) The sum of the digits of a two-digit number is $9$. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the tens digit be $x$ and the units digit be $y$. Then, $$ x+y=9 $$ Original number $$ 10x+y $$ Reversed number $$ 10y+x $$ Given, $$ 9(10x+y)=2(10y+x) $$ $$ 90x+9y=20y+2x $$ $$ 88x-11y=0 $$ $$ 8x-y=0 $$ Subtract $$ x+y=9 $$ from $$ 8x-y=0 $$ after writing $$ y=8x $$ Substitute into $$ x+y=9 $$ $$ x+8x=9 $$ $$ 9x=9 $$ $$ x=1 $$ $$ y=8 $$ Therefore, the required number is $$ 10(1)+8=18 $$

Answer:

$$ \boxed{18} $$

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(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.

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Solution:

Let the number of ₹50 notes be $x$ and the number of ₹100 notes be $y$.

Total number of notes is $25$.

$$ x+y=25 $$

Total amount is ₹2000.

$$ 50x+100y=2000 $$ Divide the second equation by $50$. $$ x+2y=40 $$ Subtract $$ x+y=25 $$ from $$ x+2y=40 $$ $$ y=15 $$ Substitute in $$ x+y=25 $$ $$ x+15=25 $$ $$ x=10 $$

Verification:

$$ 10\times50+15\times100 $$ $$ 500+1500=2000 $$

Answer:

$$ \boxed{\text{₹50 notes}=10} $$ $$ \boxed{\text{₹100 notes}=15} $$

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(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

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Solution:

Let the fixed charge be ₹$x$ and the charge for each extra day be ₹$y$.

For $7$ days, the book is kept for $4$ extra days.

$$ x+4y=27 $$

For $5$ days, the book is kept for $2$ extra days.

$$ x+2y=21 $$ Subtract $$ x+2y=21 $$ from $$ x+4y=27 $$ $$ 2y=6 $$ $$ y=3 $$ Substitute in $$ x+2y=21 $$ $$ x+2(3)=21 $$ $$ x+6=21 $$ $$ x=15 $$

Verification:

For $7$ days, $$ 15+4\times3=27 $$ For $5$ days, $$ 15+2\times3=21 $$

Answer:

$$ \boxed{\text{Fixed charge}=₹15} $$ $$ \boxed{\text{Charge for each extra day}=₹3} $$

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