Chapter 3: Pair of Linear Equations in Two Variables

Let

• $$x$$ be the number of boys,
• $$y$$ be the number of girls.

Two facts are given:

1. Total students = 10.
   \[x + y = 10 \quad(1)\]
2. Girls are 4 more than boys.
   \[y = x + 4 \;\Longrightarrow\; x - y = -4 \quad(2)\]

Graphical work

1. Line (1) : $$x + y = 10$$
   

   $$x$$	$$y$$
   0	10
   10	0

   Plot (0,10) and (10,0); join them.
2. Line (2) : $$x - y = -4$$
   

   $$x$$	$$y$$
   0	4
   4	8

   Plot (0,4) and (4,8); join them.

The two straight lines meet at the point $$ (3,7)$$ (read from the graph).

Hence

• Number of boys $$x = 3$$,
• Number of girls $$y = 7$$.

Let

• $$x$$ be the cost (in rupees) of one pencil,
• $$y$$ be the cost (in rupees) of one pen.

Two purchase statements are given:

1. 5 pencils + 7 pens cost ₹50.
   \[5x + 7y = 50 \quad(1)\]
2. 7 pencils + 5 pens cost ₹46.
   \[7x + 5y = 46 \quad(2)\]

Graphical work

1. Line (1)
   

   $$x$$	$$y$$
   0	\dfrac{50}{7}=7.14
   5	\dfrac{50-25}{7}=\dfrac{25}{7}=3.57

   Plot these two points and draw the line.
2. Line (2)
   

   $$x$$	$$y$$
   0	\dfrac{46}{5}=9.2
   2	\dfrac{46-14}{5}=\dfrac{32}{5}=6.4

   Plot these two points and draw the line.

The two lines intersect at $$ (3,5)$$.

Verification (optional algebra)

Using elimination:

Multiply (1) by 7 and (2) by 5:

$$35x + 49y = 350$$
$$35x + 25y = 230$$

Subtract ⇒ $$24y = 120 \Rightarrow y = 5$$.
Substitute in (1): $$5x + 7(5)=50 \Rightarrow 5x = 15 \Rightarrow x = 3$$.

So the intersection point observed graphically is indeed correct.

Therefore

• Cost of one pencil = ₹3,
• Cost of one pen = ₹5.

General form for each line is $$a_1x + b_1y + c_1 = 0$$ and $$a_2x + b_2y + c_2 = 0$$.

For $$5x - 4y + 8 = 0$$ and $$7x + 6y - 9 = 0$$, compare the coefficients:

$$a_1 = 5, \; b_1 = -4, \; c_1 = 8$$
$$a_2 = 7, \; b_2 = 6, \; c_2 = -9$$

Compute the three ratios:

$$\dfrac{a_1}{a_2} = \dfrac{5}{7}, \qquad \dfrac{b_1}{b_2} = \dfrac{-4}{6}= -\dfrac{2}{3}, \qquad \dfrac{c_1}{c_2} = \dfrac{8}{-9}= -\dfrac{8}{9}$$

Since $$\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$$, the pair represents intersecting lines (one unique solution).

The equations are $$9x + 3y + 12 = 0$$ and $$18x + 6y + 24 = 0$$.

Coefficients:

$$a_1 = 9, \; b_1 = 3, \; c_1 = 12$$
$$a_2 = 18, \; b_2 = 6, \; c_2 = 24$$

Ratios:

$$\dfrac{a_1}{a_2} = \dfrac{9}{18}=\dfrac{1}{2}, \qquad \dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac{1}{2}, \qquad \dfrac{c_1}{c_2}=\dfrac{12}{24}=\dfrac{1}{2}$$

All three ratios are equal, i.e. $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$, so the two equations represent coincident lines (infinitely many common points).

The equations are $$6x - 3y + 10 = 0$$ and $$2x - y + 9 = 0$$.

Coefficients:

$$a_1 = 6, \; b_1 = -3, \; c_1 = 10$$
$$a_2 = 2, \; b_2 = -1, \; c_2 = 9$$

Ratios:

$$\dfrac{a_1}{a_2}=\dfrac{6}{2}=3, \qquad \dfrac{b_1}{b_2}=\dfrac{-3}{-1}=3, \qquad \dfrac{c_1}{c_2}=\dfrac{10}{9}$$

Here $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$$ but $$\dfrac{a_1}{a_2} \ne \dfrac{c_1}{c_2}$$, therefore the pair represents parallel lines (no solution).

Write both equations in the form $$a_1x+b_1y+c_1=0$$, $$a_2x+b_2y+c_2=0$$.

Equation 1 : $$3x+2y-5=0 \Rightarrow a_1=3,\;b_1=2,\;c_1=-5$$

Equation 2 : $$2x-3y-7=0 \Rightarrow a_2=2,\;b_2=-3,\;c_2=-7$$

Find the three ratios:

$$\frac{a_1}{a_2}=\frac{3}{2},\qquad \frac{b_1}{b_2}=\frac{2}{-3},\qquad \frac{c_1}{c_2}=\frac{-5}{-7}=\frac57$$

Here $$\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$$, so the pair has one unique solution.

Hence the given equations are consistent (independent).

Standard form:

Equation 1 : $$2x-3y-8=0 \Rightarrow a_1=2,\;b_1=-3,\;c_1=-8$$

Equation 2 : $$4x-6y-9=0 \Rightarrow a_2=4,\;b_2=-6,\;c_2=-9$$

Ratios:

$$\frac{a_1}{a_2}=\frac{2}{4}=\frac12,\qquad \frac{b_1}{b_2}=\frac{-3}{-6}=\frac12,\qquad \frac{c_1}{c_2}=\frac{-8}{-9}=\frac89$$

Since $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$$, the pair has no solution.

Therefore the equations are inconsistent.

Standard form:

Equation 1 : $$\frac32x+\frac53y-7=0 \Rightarrow a_1=\frac32,\;b_1=\frac53,\;c_1=-7$$

Equation 2 : $$9x-10y-14=0 \Rightarrow a_2=9,\;b_2=-10,\;c_2=-14$$

Ratios:

$$\frac{a_1}{a_2}=\frac{\tfrac32}{9}=\frac16,\qquad \frac{b_1}{b_2}=\frac{\tfrac53}{-10}=-\frac16,\qquad \frac{c_1}{c_2}=\frac{-7}{-14}=\frac12$$

Because $$\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$$, there is exactly one solution.

The pair of equations is consistent (independent).

Standard form:

Equation 1 : $$5x-3y-11=0 \Rightarrow a_1=5,\;b_1=-3,\;c_1=-11$$

Equation 2 : $$-10x+6y+22=0 \Rightarrow a_2=-10,\;b_2=6,\;c_2=22$$

Ratios:

$$\frac{a_1}{a_2}=\frac{5}{-10}=-\frac12,\qquad \frac{b_1}{b_2}=\frac{-3}{6}=-\frac12,\qquad \frac{c_1}{c_2}=\frac{-11}{22}=-\frac12$$

Here $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$, so the lines coincide and there are infinitely many solutions.

The equations are consistent (dependent).

Standard form:

Equation 1 : $$\frac43x+2y-8=0 \Rightarrow a_1=\frac43,\;b_1=2,\;c_1=-8$$

Equation 2 : $$2x+3y-12=0 \Rightarrow a_2=2,\;b_2=3,\;c_2=-12$$

Ratios:

$$\frac{a_1}{a_2}=\frac{\tfrac43}{2}=\frac23,\qquad \frac{b_1}{b_2}=\frac{2}{3}=\frac23,\qquad \frac{c_1}{c_2}=\frac{-8}{-12}=\frac23$$

Since all three ratios are equal, the two equations represent the same straight line; thus there are infinitely many solutions.

The pair is consistent (dependent).

The given equations are

$$x+y=5\qquad\qquad(1)$$
$$2x+2y=10\qquad(2)$$

Write them in the standard form $$ax+by+c=0$$:

$$x+y-5=0\;(a_1=1,\,b_1=1,\,c_1=-5)$$
$$2x+2y-10=0\;(a_2=2,\,b_2=2,\,c_2=-10)$$

Check the ratios of the coefficients:

$$\frac{a_1}{a_2}=\frac{1}{2},\;\frac{b_1}{b_2}=\frac{1}{2},\;\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}$$

The three ratios are equal, therefore the two equations represent the same line. The pair is consistent with infinitely many solutions.

Graphical representation
To draw the line, take two convenient points from (1).
• If $$x=0$$, then $$y=5$$ ⇒ point $$(0,5)$$.
• If $$y=0$$, then $$x=5$$ ⇒ point $$(5,0)$$.
Plot these on graph paper and draw a straight line. Equation (2) gives exactly the same line, so both coincide.

Every point on the line $$x+y=5$$ (for example $$(1,4),(2,3),(4,1)\dots)$$ satisfies both equations.

The given equations are

$$x-y=8\qquad\qquad(1)$$
$$3x-3y=16\qquad(2)$$

Convert to standard form:

$$x-y-8=0\;(a_1=1,\,b_1=-1,\,c_1=-8)$$
$$3x-3y-16=0\;(a_2=3,\,b_2=-3,\,c_2=-16)$$

Coefficient ratios:

$$\frac{a_1}{a_2}=\frac{1}{3},\;\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3},\;\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}$$

Since $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$$, the lines are parallel and distinct. Hence the pair is inconsistent; there is no solution.

Graphical representation

• Equation (1): if $$x=0$$, $$y=-8$$; if $$y=0$$, $$x=8$$. Draw the line through $$(0,-8)$$ and $$(8,0)$$.
• Equation (2): divide by 3 ⇒ $$x-y=\frac{16}{3}$$.
   If $$x=0$$ ⇒ $$y=-\tfrac{16}{3}$$; if $$y=0$$ ⇒ $$x=\tfrac{16}{3}$$. Draw the line through these two points.

The two straight lines never meet, confirming the absence of any common solution.

The equations are

$$2x+y-6=0\qquad\qquad(1)$$
$$4x-2y-4=0\qquad(2)$$

Identify coefficients:

$$a_1=2,\,b_1=1,\,c_1=-6;\qquad a_2=4,\,b_2=-2,\,c_2=-4$$

Coefficient ratios:

$$\frac{a_1}{a_2}=\frac{2}{4}=\frac12,\;\frac{b_1}{b_2}=\frac{1}{-2}=-\frac12$$
Since $$\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$$, the lines intersect at a single point; the pair is consistent with a unique solution.

Finding the solution algebraically (for exact coordinates)

Rewrite (2) by dividing by 2: $$2x-y=2\qquad(2′)$$

Add (1) and (2′):
$$\bigl(2x+y\bigr)+\bigl(2x-y\bigr)=6+2\;\Rightarrow\;4x=8\;\Rightarrow\;x=2$$
Substitute in (1): $$2(2)+y=6\;\Rightarrow\;y=2$$

Thus the solution is $$(2,2)$$.

Graphical representation

• Equation (1): if $$x=0$$, $$y=6$$; if $$y=0$$, $$x=3$$. Plot $$(0,6)$$ and $$(3,0)$$ and draw the first line.
• Equation (2′): if $$x=0$$, $$y=-2$$; if $$y=0$$, $$x=1$$. Plot $$(0,-2)$$ and $$(1,0)$$ and draw the second line.

The two lines intersect exactly at the point $$(2,2)$$, verifying the calculated solution.

The equations are

$$2x-2y-2=0\qquad\qquad(1)$$
$$4x-4y-5=0\qquad(2)$$

Coefficients:

$$a_1=2,\,b_1=-2,\,c_1=-2;\qquad a_2=4,\,b_2=-4,\,c_2=-5$$

Coefficient ratios:

$$\frac{a_1}{a_2}=\frac{2}{4}=\frac12,\;\frac{b_1}{b_2}=\frac{-2}{-4}=\frac12,\;\frac{c_1}{c_2}=\frac{-2}{-5}=\frac25$$

Because $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$$, the two lines are parallel but distinct. Hence the pair is inconsistent; there is no common solution.

Graphical representation

• Equation (1): divide by 2 ⇒ $$x-y=1$$. Points: $$(0,-1)$$ and $$(1,0)$$.
• Equation (2): divide by 4 ⇒ $$x-y=\tfrac54$$. Points: $$(0,-\tfrac54)$$ and $$(\tfrac54,0)$$.

Both are parallel straight lines that never meet, confirming that the system has no solution.

The given line is $$2x + 3y - 8 = 0$$   … (1)

Let the required second line be $$a_2x + b_2y + c_2 = 0$$   … (2)

For (1) and (2) to represent intersecting lines we need their slopes to be different, i.e.

$$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$

Choosing small integral coefficients that violate this equality, take

$$x - y + 1 = 0$$

Here $$a_2 = 1,\; b_2 = -1,\; c_2 = 1$$.

Now $$\frac{a_1}{a_2}=\frac{2}{1}=2$$ while $$\frac{b_1}{b_2}=\frac{3}{-1}=-3$$, clearly $$2\neq -3$$. Hence the two lines intersect at one unique point.

The given line is again (1) $$2x + 3y - 8 = 0$$.

For the second line to be parallel (but not coincident) with (1) we require

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}\quad\text{and}\quad \frac{c_1}{c_2}\neq\frac{a_1}{a_2}$$

Multiplying every coefficient of (1) by 2 would give an identical ratio for $$c$$ as well, so we alter only the constant term:

Choose $$4x + 6y - 9 = 0$$.

Now

$$\frac{a_1}{a_2}=\frac{2}{4}=\tfrac12, \qquad \frac{b_1}{b_2}=\frac{3}{6}=\tfrac12$$

but

$$\frac{c_1}{c_2}=\frac{-8}{-9}=\tfrac89\neq\tfrac12$$.

Because the first two ratios are equal while the third is different, the two lines are parallel and distinct.

For the required line to be coincident with $$2x + 3y - 8 = 0$$ we must have

$$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$.

This is achieved simply by multiplying the whole equation by the same non-zero constant, say 2:

$$4x + 6y - 16 = 0$$.

Indeed, $$\frac{2}{4}=\frac{3}{6}=\frac{-8}{-16}=\tfrac12$$, so every ratio is equal and both equations describe the same straight line.

The equations are

$$x + y = 14 \quad (1)$$

$$x - y = 4 \quad (2)$$

From (1) express $$x$$:

$$x = 14 - y$$

Substitute this in (2):

$$14 - y - y = 4$$

$$14 - 2y = 4$$

$$-2y = -10$$

$$y = 5$$

Put $$y = 5$$ in $$x = 14 - y$$:

$$x = 14 - 5 = 9$$

Hence $$x = 9,\; y = 5$$.

The equations are

$$s - t = 3 \quad (1)$$

$$\dfrac{s}{3} + \dfrac{t}{2} = 6 \quad (2)$$

From (1): $$s = t + 3$$.

Substitute in (2):

$$\dfrac{t + 3}{3} + \dfrac{t}{2} = 6$$

Take LCM 6:

$$\dfrac{2(t + 3) + 3t}{6} = 6$$

$$5t + 6 = 36$$

$$5t = 30$$

$$t = 6$$

Then $$s = t + 3 = 9$$.

Thus $$s = 9,\; t = 6$$.

The equations are

$$3x - y = 3 \quad (1)$$

$$9x - 3y = 9 \quad (2)$$

From (1): $$y = 3x - 3$$.

Substitute in (2):

$$9x - 3(3x - 3) = 9$$

$$9x - 9x + 9 = 9$$

$$9 = 9$$

The identity is true for every $$x$$; hence the two equations are dependent and have infinitely many solutions.

If $$x = k$$ (any real number), then $$y = 3k - 3$$.

Solution set: $$\{(k,\;3k - 3)\;|\;k \in \mathbb{R}\}$$.

Given

$$0.2x + 0.3y = 1.3 \quad (1)$$

$$0.4x + 0.5y = 2.3 \quad (2)$$

Multiply each equation by 10:

$$2x + 3y = 13 \quad (3)$$

$$4x + 5y = 23 \quad (4)$$

From (3): $$2x = 13 - 3y \;\Rightarrow\; x = \dfrac{13 - 3y}{2}$$.

Substitute in (4):

$$4\left(\dfrac{13 - 3y}{2}\right) + 5y = 23$$

$$2(13 - 3y) + 5y = 23$$

$$26 - 6y + 5y = 23$$

$$26 - y = 23$$

$$y = 3$$

Then $$x = \dfrac{13 - 3\times3}{2} = 2$$.

So $$x = 2,\; y = 3$$.

The equations are

$$\sqrt{2}\,x + \sqrt{3}\,y = 0 \quad (1)$$

$$\sqrt{3}\,x - \sqrt{8}\,y = 0 \quad (2)$$

From (1): $$\sqrt{2}\,x = -\sqrt{3}\,y \;\Rightarrow\; x = -\dfrac{\sqrt{3}}{\sqrt{2}}\,y$$.

Substitute in (2):

$$\sqrt{3}\left(-\dfrac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}\,y = 0$$

$$-\dfrac{3}{\sqrt{2}}y - \sqrt{8}\,y = 0$$

Take $$y$$ common:

$$y\left(-\dfrac{3}{\sqrt{2}} - \sqrt{8}\right) = 0$$

The bracketed coefficient is non–zero, hence $$y = 0$$.

Putting $$y = 0$$ in (1): $$\sqrt{2}\,x = 0 \;\Rightarrow\; x = 0$$.

Thus $$x = 0,\; y = 0$$.

The equations are

$$\dfrac{3x}{2} - \dfrac{5y}{3} = -2 \quad (1)$$

$$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} \quad (2)$$

From (2):

$$\dfrac{x}{3} = \dfrac{13}{6} - \dfrac{y}{2}$$

$$x = 3\left(\dfrac{13}{6} - \dfrac{y}{2}\right) = \dfrac{13 - 3y}{2}$$

Substitute this $$x$$ in (1):

$$\dfrac{3}{2}\left(\dfrac{13 - 3y}{2}\right) - \dfrac{5y}{3} = -2$$

$$\dfrac{39 - 9y}{4} - \dfrac{5y}{3} = -2$$

Take LCM 12:

$$\dfrac{117 - 27y - 20y}{12} = -2$$

$$\dfrac{117 - 47y}{12} = -2$$

$$117 - 47y = -24$$

$$-47y = -141$$

$$y = 3$$

Put $$y = 3$$ in $$x = \dfrac{13 - 3y}{2}$$:

$$x = \dfrac{13 - 9}{2} = 2$$

Hence $$x = 2,\; y = 3$$.

Let the two numbers be $$x$$ and $$y$$ (take $$x$$ as the larger one).

According to the conditions:

• Difference is 26    $$x - y = 26$$
• One number is three times the other    $$x = 3y$$

Substitute $$x = 3y$$ in $$x - y = 26$$:

$$3y - y = 26 \ 2y = 26 \ y = 13$$

Put $$y = 13$$ in $$x = 3y$$:

$$x = 3(13) = 39$$

Let the larger angle be $$L$$ and the smaller angle be $$S$$.

• Supplementary    $$L + S = 180$$
• Larger exceeds smaller by 18°    $$L - S = 18$$

From the first equation, $$L = 180 - S$$.

Substitute in the second:

$$180 - S - S = 18 \ 180 - 2S = 18 \ 2S = 162 \ S = 81$$

Then $$L = 180 - 81 = 99$$

Let the cost of one bat be $$x$$ rupees and that of one ball be $$y$$ rupees.

Forming equations:

• $$7x + 6y = 3800$$
• $$3x + 5y = 1750$$

From the second: $$3x = 1750 - 5y \ \Rightarrow \ x = \dfrac{1750 - 5y}{3}$$

Substitute in the first:

$$7\left(\dfrac{1750 - 5y}{3}\right) + 6y = 3800$$

$$\dfrac{12250 - 35y}{3} + 6y = 3800$$

Multiply by 3:

$$12250 - 35y + 18y = 11400$$

$$12250 - 17y = 11400 \;\Rightarrow\; -17y = -850 \;\Rightarrow\; y = 50$$

Now $$x = \dfrac{1750 - 5(50)}{3} = \dfrac{1750 - 250}{3} = \dfrac{1500}{3} = 500$$

Let the fixed charge be $$x$$ rupees and the charge per km be $$y$$ rupees.

• For 10 km: $$x + 10y = 105$$
• For 15 km: $$x + 15y = 155$$

Subtract the first from the second:

$$5y = 50 \;\Rightarrow\; y = 10$$

Substitute in $$x + 10y = 105$$:

$$x + 10(10) = 105 \;\Rightarrow\; x = 5$$

For 25 km, charge = $$x + 25y = 5 + 25(10) = 255$$

Let the required fraction be $$\dfrac{x}{y} \; (y \neq 0).$$

• If 2 is added: $$\dfrac{x+2}{y+2} = \dfrac{9}{11} \;\Rightarrow\; 11(x+2) = 9(y+2)$$
  $$11x + 22 = 9y + 18 \;\Rightarrow\; 11x - 9y = -4 \;(1)$$
• If 3 is added: $$\dfrac{x+3}{y+3} = \dfrac{5}{6} \;\Rightarrow\; 6(x+3) = 5(y+3)$$
  $$6x + 18 = 5y + 15 \;\Rightarrow\; 6x - 5y = -3 \;(2)$$

From (2): $$6x = 5y - 3 \;\Rightarrow\; x = \dfrac{5y - 3}{6}$$

Substitute in (1):

$$11\left(\dfrac{5y - 3}{6}\right) - 9y = -4$$

$$\dfrac{55y - 33}{6} - 9y = -4$$

Converting $$9y$$ to sixths: $$\dfrac{54y}{6}$$

$$\dfrac{55y - 33 - 54y}{6} = -4 \;\Rightarrow\; \dfrac{y - 33}{6} = -4$$

$$y - 33 = -24 \;\Rightarrow\; y = 9$$

Then $$x = \dfrac{5(9) - 3}{6} = \dfrac{45 - 3}{6} = \dfrac{42}{6} = 7$$

Let Jacob's present age be $$J$$ years and his son's present age be $$S$$ years.

• Five years hence: $$J + 5 = 3(S + 5)$$
• Five years ago: $$J - 5 = 7(S - 5)$$

Simplify:

$$J + 5 = 3S + 15 \;\Rightarrow\; J - 3S = 10 \;(1)$$

$$J - 5 = 7S - 35 \;\Rightarrow\; J - 7S = -30 \;(2)$$

Subtract (2) from (1):

$$(J - 3S) - (J - 7S) = 10 - (-30)$$
$$-3S + 7S = 40 \;\Rightarrow\; 4S = 40 \;\Rightarrow\; S = 10$$

Put $$S = 10$$ in (1):

$$J - 3(10) = 10 \;\Rightarrow\; J = 40$$

Elimination method

Equation (1) $$x + y = 5$$
Equation (2) $$2x - 3y = 4$$

Multiply (1) by 3 so that the coefficients of $$y$$ become opposites:

$$3x + 3y = 15 \\[6pt] \text{(3)}$$

Add (2) and (3):

$$\begin{aligned} (2)&:&\;2x - 3y &= 4\\ (3)&:&\;3x + 3y &= 15\\ \hline 5x &= 19 \end{aligned} \Rightarrow x = \dfrac{19}{5}$$

Put $$x = \dfrac{19}{5}$$ in (1):

$$\dfrac{19}{5} + y = 5 \;\Rightarrow\; y = 5 - \dfrac{19}{5} = \dfrac{6}{5}$$

Substitution method

From (1) $$y = 5 - x$$. Substitute in (2):

$$2x - 3(5 - x) = 4 \;\Rightarrow\; 2x - 15 + 3x = 4 \;\Rightarrow\; 5x = 19$$

$$x = \dfrac{19}{5},\; y = 5 - \dfrac{19}{5} = \dfrac{6}{5}$$

Thus $$x = \dfrac{19}{5},\; y = \dfrac{6}{5}$$.

Given equations

(1) $$3x + 4y = 10$$
(2) $$2x - 2y = 2$$

Divide (2) by 2 to simplify:

(3) $$x - y = 1$$

Elimination method

Multiply (3) by 4 to match the coefficient of $$y$$ in (1):

$$4x - 4y = 4 \;\;\;(4)$$

Add (1) and (4):

$$\begin{aligned} 3x + 4y &= 10\\ 4x - 4y &= 4\\ \hline 7x &= 14 \end{aligned} \Rightarrow x = 2$$

Substitute $$x = 2$$ in (3): $$2 - y = 1 \;\Rightarrow\; y = 1$$

Substitution method

From (3) $$y = x - 1$$. Substitute in (1):

$$3x + 4(x - 1) = 10 \;\Rightarrow\; 3x + 4x - 4 = 10 \;\Rightarrow\; 7x = 14 \;\Rightarrow\; x = 2$$

Hence $$y = 2 - 1 = 1$$.

Therefore $$x = 2,\; y = 1$$.

Rewrite the equations in standard form

(1) $$3x - 5y - 4 = 0 \;\Rightarrow\; 3x - 5y = 4$$
(2) $$9x = 2y + 7 \;\Rightarrow\; 9x - 2y = 7$$

Elimination method

Multiply (1) by 2 and (2) by 5 to make the coefficients of $$y$$ equal:

$$\begin{aligned} 6x - 10y &= 8 \quad\;\;(3)\\ 45x - 10y &= 35 \quad (4) \end{aligned}$$

Subtract (3) from (4):

$$45x - 10y - (6x - 10y) = 35 - 8 \;\Rightarrow\; 39x = 27 \;\Rightarrow\; x = \dfrac{27}{39} = \dfrac{9}{13}$$

Substitute $$x = \dfrac{9}{13}$$ in (1):

$$3\left(\dfrac{9}{13}\right) - 5y = 4 \;\Rightarrow\; \dfrac{27}{13} - 5y = 4$$

$$-5y = 4 - \dfrac{27}{13} = \dfrac{25}{13} \;\Rightarrow\; y = -\dfrac{25}{13}\times\dfrac{1}{5} = -\dfrac{5}{13}$$

Substitution method

From (2): $$9x = 2y + 7 \;\Rightarrow\; y = \dfrac{9x - 7}{2}$$

Substitute in (1):

$$3x - 5\left(\dfrac{9x - 7}{2}\right) = 4$$

$$\Rightarrow\; 3x - \dfrac{45x - 35}{2} = 4$$

Multiply by 2: $$6x - 45x + 35 = 8 \;\Rightarrow\; -39x = -27 \;\Rightarrow\; x = \dfrac{9}{13}$$

Then $$y = \dfrac{9(9/13) - 7}{2} = \dfrac{81/13 - 91/13}{2} = -\dfrac{10/13}{2} = -\dfrac{5}{13}$$

Thus $$x = \dfrac{9}{13},\; y = -\dfrac{5}{13}$$.

Convert both equations to integral coefficients

(1) $$\dfrac{x}{2} + \dfrac{2y}{3} = -1 \;\Rightarrow\; 6\Bigl(\dfrac{x}{2} + \dfrac{2y}{3}\Bigr) = 6(-1) \;\Rightarrow\; 3x + 4y = -6$$
(2) $$x - \dfrac{y}{3} = 3 \;\Rightarrow\; 3\Bigl(x - \dfrac{y}{3}\Bigr) = 3\times 3 \;\Rightarrow\; 3x - y = 9$$

Elimination method

Equations now are
(1′) $$3x + 4y = -6$$
(2′) $$3x - y = 9$$

Subtract (2′) from (1′):

$$5y = -6 - 9 = -15 \;\Rightarrow\; y = -3$$

Substitute $$y = -3$$ in (2′):

$$3x - (-3) = 9 \;\Rightarrow\; 3x + 3 = 9 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$$

Substitution method

From (2′): $$3x - y = 9 \;\Rightarrow\; y = 3x - 9$$

Put this in (1′):

$$3x + 4(3x - 9) = -6 \;\Rightarrow\; 3x + 12x - 36 = -6$$

$$15x = 30 \;\Rightarrow\; x = 2,\; y = 3(2) - 9 = -3$$

Therefore $$x = 2,\; y = -3$$.

Let the required fraction be $$\dfrac{x}{y}$$, where $$x$$ is the numerator and $$y$$ is the denominator.

According to the first condition,

$$\dfrac{x+1}{y-1}=1 \;\Longrightarrow\; x+1=y-1 \;\Longrightarrow\; x-y=-2 \;\;(1)$$

According to the second condition,

$$\dfrac{x}{y+1}=\dfrac12 \;\Longrightarrow\; 2x=y+1 \;\Longrightarrow\; 2x-y=1 \;\;(2)$$

We now solve the pair of linear equations (1) and (2) by the elimination method.

Subtract (1) from (2):
$$\bigl(2x-y\bigr)-\bigl(x-y\bigr)=1-(-2)$$
$$2x-y-x+y=x+0y=3$$
$$\therefore\; x=3$$

Substitute $$x=3$$ in (1):
$$3-y=-2\;\Longrightarrow\; y=5$$

Hence the fraction is $$\dfrac{3}{5}$$.

Let the present ages of Nuri and Sonu be $$x$$ years and $$y$$ years respectively.

Five years ago:
$$x-5=3(y-5)\;\Longrightarrow\; x-5=3y-15\;\Longrightarrow\; x-3y=-10\;\;(1)$$

Ten years hence:
$$x+10=2(y+10)\;\Longrightarrow\; x+10=2y+20\;\Longrightarrow\; x-2y=10\;\;(2)$$

Eliminate $$x$$ by subtracting (1) from (2):
$$\{x-2y\}-\{x-3y\}=10-(-10)$$
$$-2y+3y=y=20$$

Put $$y=20$$ in (2):
$$x-2(20)=10\;\Longrightarrow\; x-40=10\;\Longrightarrow\; x=50$$

Therefore, Nuri is $$50$$ years old and Sonu is $$20$$ years old.

Let the tens digit be $$x$$ and the units digit be $$y$$.
Original number = $$10x+y$$.
Number with digits reversed = $$10y+x$$.

Given, the sum of the digits is 9:
$$x+y=9\;\;(1)$$

Also, nine times the number equals twice the reversed number:
$$9(10x+y)=2(10y+x)$$
$$90x+9y=20y+2x$$
$$88x=11y\;\Longrightarrow\; 8x=y\;\;(2)$$

Substitute $$y=8x$$ in (1):
$$x+8x=9\;\Longrightarrow\; 9x=9\;\Longrightarrow\; x=1$$

Then $$y=8$$. Hence the required number is
$$10x+y=10(1)+8=18$$.

Let Meena receive $$x$$ notes of $$\text{₹}\,\,50$$ and $$y$$ notes of $$\text{₹}\,\,100$$.

Total number of notes:
$$x+y=25\;\;(1)$$

Total amount:
$$50x+100y=2000\;\Longrightarrow\; x+2y=40\;\;(2)$$

Subtract (1) from (2):
$$\bigl(x+2y\bigr)-\bigl(x+y\bigr)=40-25$$
$$y=15$$

Substitute $$y=15$$ in (1):
$$x+15=25\;\Longrightarrow\; x=10$$

Therefore, Meena received 10 notes of $$\text{₹}\,\,50$$ and 15 notes of $$\text{₹}\,\,100$$.

Let the fixed charge for the first three days be $$\text{₹}\,\,x$$ and the additional charge per extra day be $$\text{₹}\,\,y$$.

Saritha: kept the book 7 days.
Total charge: $$x+4y=27$$   (since 7−3 = 4 extra days)   $$\;(1)$$

Susy: kept the book 5 days.
Total charge: $$x+2y=21$$   (5−3 = 2 extra days)   $$\;(2)$$

Subtract (2) from (1):
$$\{x+4y\}-\{x+2y\}=27-21$$
$$2y=6\;\Longrightarrow\; y=3$$

Put $$y=3$$ in (2):
$$x+2(3)=21\;\Longrightarrow\; x+6=21\;\Longrightarrow\; x=15$$

Hence the fixed charge is $$\text{₹}\,\,15$$ and the additional charge per extra day is $$\text{₹}\,\,3$$.