Algebra is one subject where you can never have enough practice. The more you work out problems, the easier they become. You won’t find much time or opportunity to practice in the classroom, where the focus is on teaching new methods and covering the topics in the curriculum. You will have to make time to work out problems. But how do you know if your work is correct? This is where free algebra problem solvers come in. Free online algebra solvers provide students with practice problems and worksheets and also check your work and correct any mistakes. Most of the free online algebra solvers or algebra free solvers have plenty of practice material and also explain any bits that you don’t know. The explanations are simple and straight forward and are easily understood by students of all grades.

Given (3x - y)^{2} - 12(x + 4y) + 23

Here polynomial is divided into three parts so it is a trinomial.

Step 1:

Solve each part of the polynomial

**First part**

(3x - y)^{2} = (3x)^{2} + y^{2} - 2 * 3x * y

= 9x^{2} + y^{2} - 6xy

[(a - b)^{2} = a^{2} + b^{2} - 2ab]

Second part

12(x + 4y) = 12 * x + 12 * 4y

= 12x + 48y

Third part

23 = 23

**Step 2:**

=> (3x - y)^{2} - 12(x + 4y) + 23 = 9x^{2} + y^{2} - 6xy - (12x + 48y) + 23

= 9x^{2} + y^{2} - 6xy - 12x - 48y + 23

=> (3x - y)^{2} - 12(x + 4y) + 23 = 9x^{2} + y^{2} - 6xy - 12x - 48y + 23. **answer**

Here polynomial is divided into three parts so it is a trinomial.

Step 1:

Solve each part of the polynomial

(3x - y)

= 9x

[(a - b)

Second part

12(x + 4y) = 12 * x + 12 * 4y

= 12x + 48y

Third part

23 = 23

= 9x

=> (3x - y)

Given a - 16a^{-3}

Here common factor is 'a'.

Step 1:

When factoring the common factor, take the lowest power of the factor.

Lowest power of the factor = a^{-3}

=> a - 16a^{-3} = a^{-3} ($\frac{a}{a^{-3}}$ - 16)

= a^{-3} ($\frac{a^1}{a^{-3}}$ - 16)

= a^{-3} ($a^{1 - (-3)}$ - 16)

[$\frac{a^m}{a^n}$ = $a^{m - n}$]

= a^{-3} ($a^{1 + 3}$ - 16)

= a^{-3} (a^{4 }- 16)

**Step 2:**

To eliminate the negative exponent, convert it into a fraction.

=> a^{-3} (a^{4 }- 16) = $\frac{1}{a^{3}}$ (a^{4} - 16)

= $\frac{1}{a^{3}}$ (a^{4} - 2^{4})

= $\frac{1}{a^{3}}$ ($(a^2)^2 - (2^2)^2$)

[a^{2} - b^{2} = (a - b)(a + b)]

= $\frac{1}{a^{3}}$(a^{2} - b^{2} )(a^{2} + b^{2} )

= $\frac{1}{a^{3}}$(a - b)(a + b)(a^{2} + b^{2} )

=> a^{-3} (a^{4 }- 16) = $\frac{1}{a^{3}}$(a - b)(a + b)(a^{2} + b^{2} )

Here common factor is 'a'.

Step 1:

When factoring the common factor, take the lowest power of the factor.

Lowest power of the factor = a

=> a - 16a

= a

= a

[$\frac{a^m}{a^n}$ = $a^{m - n}$]

= a

= a

To eliminate the negative exponent, convert it into a fraction.

=> a

= $\frac{1}{a^{3}}$ (a

= $\frac{1}{a^{3}}$ ($(a^2)^2 - (2^2)^2$)

[a

= $\frac{1}{a^{3}}$(a

= $\frac{1}{a^{3}}$(a - b)(a + b)(a

=> a

Given equation of lines are x + 2y = 3 and x + y = 7

**Step 1:**

Since the point of intersection lie on the two lines, its coordinates x and y satisfy the two equations simultaneously.

To find the point of intersection, find the values of x and y

x + 2y = 3 ................................(1)

x + y = 7 ..................................(2)

**Step 2:**

Eliminate y from equation (1), choose equation (2)

=> y = 7 - x .....................................(3)

(3) put in (1)

=> x + 2(7 - x) = 3

=> x + 14 - 2x = 3

=> - x + 14 = 3

=> - x = 3 - 14

=> - x = - 11

=> x = 11

**Step 3:**

To find the value of y put x = 11 in equation (3).

=> y = 7 - 11

=> y = - 4

Hence the point of intersection of the lines is (x, y) = (11, - 4).**answer**

Since the point of intersection lie on the two lines, its coordinates x and y satisfy the two equations simultaneously.

To find the point of intersection, find the values of x and y

x + 2y = 3 ................................(1)

x + y = 7 ..................................(2)

=> y = 7 - x .....................................(3)

(3) put in (1)

=> x + 2(7 - x) = 3

=> x + 14 - 2x = 3

=> - x + 14 = 3

=> - x = 3 - 14

=> - x = - 11

=> x = 11

=> y = 7 - 11

=> y = - 4

Hence the point of intersection of the lines is (x, y) = (11, - 4).