The highest power of the term in the given expression is called degree. The expression having degree one is a linear expression. So, 4x – 3y is a linear expression.

The two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 will be parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. ^_$\because$        $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$ and $\frac{c_1}{c_2} = \frac{5}{12} $. ^{_$\therefore$}        $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ Hence, 2x + 3y = 5 and 4x + 6y = 12 are parallel lines.

2x – y + 4 = 2 (0) – (3) + 4 = 0 – 3 + 4 = 1

$\frac{2x + 4}{x+ 5} = 7$             2x + 4 = 7 (x + 5)             2x + 4 = 7x + 35             2x – 7x = 35 – 4             – 5x = 31 ^{_$\therefore$}        x =$\frac{-31}{5}$

These two lines N and M do not intersect each other. $\therefore$ There is no common solution.

The highest power of the term in the given equation is called degree. An equation having degree one, is called a linear equation. ^_$\because$x = 4 is an equation of degree one, therefore it is a linear equation.

11(x + 7) = 2x + 13             11x + 77 = 2x + 13             11x – 2x = 13 – 77             9x = – 64             x =$\frac{-64}{9}$.

Let Jatin has y number of pens and Lalit has x number of pens. Jatin has 4 pens more than twice the number of pens Lalit has. y = 4 + 2x

$\frac{7x^2 + 2x + 5}{x+5}$ = $\frac{7(-1)^2 + 2(-1) + 5}{(-1)+5} = \frac{7-2 + 5}{4}$= $\frac{5}{2}$.

The highest power of the term in the given equation is called degree. The equation having a degree one is called a linear equation. Therefore, 7x + 10 = 0 is the linear equation.

7x + 2 = 4x + 5 7x – 4x = 5 – 2 3x = 3 x = 1

$\frac{7x^2y + 2xy^2 + 3}{(x+4)^3}$ = $\frac{7(1)^2 2 + 2(1)(2)^2 + 3}{(1+4)^3}$= $\frac{14+8+3}{125} = \frac{1}{5}$ 

When 2 lines coincide, every point is common, then there are infinite solutions.

x + y – 3             = 1 + 2 – 3             = 3 – 3 = 0

The highest power of the term in the given expression is called degree. An expression having degree one is a linear expression. Therefore, 4x + 3y + 2 is the linear expression.

For the pair of equations to represent coincident lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ In lines x + 2y = 7 and 2x + 4y = 14, $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} =\frac{2}{4} =\frac{1}{2}, \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$ So, _{$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$}

size = 2 > 4 (3x + 2) = 7x – 3 12x + 8 = 7x – 3 12x – 7x = – 3 – 8 5x = – 11 x = $\frac{-11}{5}$

Let length of the rectangle be y and breadth of the rectangle be x.

As length of the rectangle is 6 units more than twice its breadth, $\therefore$ y = 6 + 2x 

The highest power of the term in the given expression is called degree. The expression having a degree one is a linear expression. Therefore, x + z is a linear expression.

4x³ - 2x + 3 = 4 ( 1)³ - 2( 1) + 3 = 4 (1) - 2 + 3 = 4 - 2 + 3 = 5

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel when $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. In lines x – y = 7 and 2x – 2y = 15, $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} =\frac{-1}{-2} , \frac{c_1}{c_2} = \frac{7}{15}$ ^{_$\therefore$}        $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\frac{x+4}{x+2} = \frac{1}{5}$ 5 (x + 4) = x + 2 5x + 20 = x + 2 5x – x = 2 – 20 4x = –18 x = $\frac{-18}{4}$ x =$\frac{-9}{2}$

Runs scored by Sachin is 25 more than thrice the runs score by Rahul.

8x + 9 = 3x + 2 5x = – 9 + 2 ^{X = - 7/5}  

Number of red chairs is 5 less than twice the number of yellow ones, 2x = y + 5

x + 2y - 4 = (1) + 2(1) - 4 = 1 + 2 - 4 = - 1

A pair of lines is coincident, if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. In lines, x + 2y = 3 and 4x + 8y = 12 $\frac{a_1}{a_2} = \frac{1}{4}, \frac{b_1}{b_2} =\frac{2}{8} = \frac{1}{4} , \frac{c_1}{c_2} = \frac{3}{12} = \frac{1}{4}$ $\therefore$   $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

The highest power of the term in the given equation is called degree. The equation having a degree one is called a linear equation.

$\frac{x + 3}{x- 3} = 4$ x + 3 = 4 (x – 3) x + 3 = 4x – 12 x – 4x = – 12 – 3 – 3x = – 15 x = $\frac{-15}{3}$ x = 5

x is the length and y is the breadth. x is equal to 3 more than the twice of y. Hence x = 2y + 3.

2x - 2y + 4 = 2(4) - 2(0) + 4 = 8 - 0 + 4 = 8 + 4 = 12

When 2 lines intersect at one point, the pair of linear equations has a unique solution.

Denominator is 2 more than thrice its numerator, y = 2 + 3x

$\frac{2x + 5}{x- 5} = \frac{1}{8}$ 16x + 40 = x – 5 16x – x = – 40 – 5 15x = – 45 x = – $\frac{45}{15}$= – 3

The highest power of the term in the given equation is called degree. The equation having a degree one is called a linear equation. Therefore, the equation x + y - 3 = 0 is a linear equation.

$\frac{8(x)^3 + 3(x)^2y + 2y}{x+y}$ = $\frac{8(-1)^3 + 3(-1)^2y + 2y}{-1+y}$ = $\frac{8(-1)^3 + 3(-1)^2 2 + 2(2)}{-1+2}$ = $\frac{8(-1) + 6 + 4}{1}$ = – 8 + 6 + 4 = 2

Denominator is 5 more than twice its numerator, y = 5 + 2x

For two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be parallel, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. In lines x – y = 17 and 2x – 2y = 38, $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} =\frac{-1}{-2} = \frac{1}{2} , \frac{c_1}{c_2} = \frac{3}{12} = \frac{17}{38}$ Therefore, ^{_{$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$}} Hence, the lines are parallel.

The expression having a degree one is called a linear expression. Therefore, 3xy + 7 is not a linear expression because its degree is 2.