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Step 1:

Find the line passes through the points (2, 4) and (5, 8).

Let (x_{1}, y_{1}) = (2, 4) and (x_{2} , y_{2} ) = (5, 8)

[Equation of line passes through a point is:

y - y_{1} = m (x - x_{1}) Where 'm' is the slope of the line]

m = $\frac{y_2 - y_1}{x_2 - x_1}$

=> m = $\frac{8 - 4}{5 - 2}$

=> m = $\frac{4}{3}$

Step 2:

Equation of line passes through point (2, 4)

=> y - 4 = $\frac{4}{3}$(x - 2)

=> 3(y - 4) = 4(x - 2)

=> 3y - 12 = 4x - 8

=> 4x - 3y + 4 = 0

Which is required equation of line.

Find the line passes through the points (2, 4) and (5, 8).

Let (x

[Equation of line passes through a point is:

y - y

m = $\frac{y_2 - y_1}{x_2 - x_1}$

=> m = $\frac{8 - 4}{5 - 2}$

=> m = $\frac{4}{3}$

Step 2:

Equation of line passes through point (2, 4)

=> y - 4 = $\frac{4}{3}$(x - 2)

=> 3(y - 4) = 4(x - 2)

=> 3y - 12 = 4x - 8

=> 4x - 3y + 4 = 0

Which is required equation of line.

Given, (2x - 3)(4x^{2} - 5x + 6)

=> 2x (4x^{2} - 5x + 6) - 3(4x^{2} - 5x + 6)

=> 2x * 4x^{2} - 2x * 5x + 2x * 6 - 3 * 4x^{2} + 3 * 5x - 3 * 6

=> 8x^{3} - 10x^{2} + 12x - 12x^{2} +15x - 18

=> 8x^{3} - 22x^{2} + 27x - 18

=> 2x (4x

=> 2x * 4x

=> 8x

=> 8x

y - x = -7 and 2x

Given system of equations

y - x = -7 and 2x^{2 }+ y = - 4

Step 1:

y - x = -7 .........................(1)

2x^{2 }+ y = - 4 ...........................(2)

(1) => y = -7 + x

Step 2:

Put y = -7 + x in equation (2)

=> 2x^{2} + (-7 + x) = -4

=> 2x^{2} - 7 + x = - 4

=> 2x^{2} + x - 3 = 0

which is quadratic equation

Step 3:

Solve for x,

2x^{2} + x - 3 = 0

=> 2x^{2} - 2x + 3x - 3 = 0

=> 2x(x - 1) + 3(x - 1) = 0

=> (2x + 3)(x - 1) = 0

either 2x + 3 = 0 or x - 1 = 0

=> 2x + 3 = 0 =>**x =** $\frac{-3}{2}$

and x - 1 = 0 =>** x = 1**

Step 4:

When x = $\frac{-3}{2}$, Put in equation (1)

=> y - $\frac{-3}{2}$ = -7

=> y = - 7 + $\frac{-3}{2}$

=> y = $\frac{-17}{2}$

When**x = 1**, Put in equation (1)

=> y - 1 = - 7

=> y = - 6

Hence solution of the system is ($\frac{-3}{2}$, $\frac{-17}{2}$) or (1, - 6).

y - x = -7 and 2x

Step 1:

y - x = -7 .........................(1)

2x

(1) => y = -7 + x

Step 2:

Put y = -7 + x in equation (2)

=> 2x

=> 2x

=> 2x

which is quadratic equation

Step 3:

Solve for x,

2x

=> 2x

=> 2x(x - 1) + 3(x - 1) = 0

=> (2x + 3)(x - 1) = 0

either 2x + 3 = 0 or x - 1 = 0

=> 2x + 3 = 0 =>

and x - 1 = 0 =>

Step 4:

When x = $\frac{-3}{2}$, Put in equation (1)

=> y - $\frac{-3}{2}$ = -7

=> y = - 7 + $\frac{-3}{2}$

=> y = $\frac{-17}{2}$

When

=> y - 1 = - 7

=> y = - 6

Hence solution of the system is ($\frac{-3}{2}$, $\frac{-17}{2}$) or (1, - 6).