College level math problems require a different approach from the way students are used to solving high school math. The topics are definitely more detailed and complex and students are expected to keep up with long classes which never seem to end. The good news for all the college students looking for a way out of their mathematics woes is that online help is available for college level math problems, anywhere, anytime. College algebra math problems can be especially difficult. One of the best ways to improve is to practice college math problems online. Math help websites have a repository of college math problems and answers which students can keep solving and figure out how to tackle different problems. College math word problems cover a variety of topics and use familiar situations and scenarios that aid the learning process.

## College Level Math Problems

Virtual learning is as effective as classroom teaching and the flexibility it gives students in terms of time and place is a huge advantage. College math practice problems cover all types of questions including college algebra math problems. Online math solvers provide flexibility and ease which students increasingly find to be more adaptable to their own study habits and patterns.

## Solved Examples

**Question 1:**Find the equation of line which passes through the points (2, 4) and (5, 8).

**Solution:**

Step 1:

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.

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.

**Question 2:**Solve (2x - 3)(4x

^{2}- 5x + 6)

**Solution:**

Given, (2x - 3)(4x

=> 2x (4x

=> 2x * 4x

=> 8x

=> 8x

^{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**Question 3:**Solve

y - x = -7 and 2x

^{2 }+ y = - 4

**Solution:**

Given system of equations

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).

y - x = -7 and 2x

^{2 }+ y = - 4Step 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 = 0which 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).