Algebra formulas for class 10 are provided here for students. The study of Algebra requires an intrinsic and deep understanding of formulas, terms and concept. Hence, to ease this task, we provide to you “Algebra Formulas for class 10”, a thorough and complete guide exclusively drafted to boost the confidence of the students. All algebra formulas and algebraic identities for class 10 in a single page format will help students to grasp and understand every concept thoroughly and solve problems easily.
The CBSE class 10 Mathematics consists of algebra in many chapters like:
- Algebraic method of solving a pair of linear equation- Elimination Method
- Algebraic method of solving a pair of linear equation- Substitution Method
- Algebraic method of solving a pair of the linear equation- Cross Multiplication Method
List of Algebra Formulas for Class 10
All the algebraic identities for class 10 formulas are provided here. A student may feel difficult to note them and read. To make it easy for them, all the important algebra formulas for class 10 are listed below.
| Algebraic Identities For Class 10 |
\(\begin{array}{l}\mathbf{(a+b)^{2}}=a^2+2ab+b^{2}\end{array} \) |
\(\begin{array}{l}\mathbf{(a-b)^{2}}=a^{2}-2ab+b^{2}\end{array} \) |
\(\begin{array}{l}\mathbf{\left (a + b \right ) \left (a – b \right ) } = a^{2} – b^{2}\end{array} \) |
\(\begin{array}{l}\mathbf{ \left (x + a \right )\left (x + b \right ) }= x^{2} + \left (a + b \right )x + ab\end{array} \) |
\(\begin{array}{l}\mathbf{\left (x + a \right )\left (x – b \right ) }= x^{2} + \left (a – b \right )x – ab\end{array} \) |
\(\begin{array}{l}\mathbf{\left (x – a \right )\left (x + b \right )}= x^{2} + \left (b – a \right )x – ab\end{array} \) |
\(\begin{array}{l}\mathbf{\left (x – a \right )\left (x – b \right ) }= x^{2} – \left (a + b \right )x + ab\end{array} \) |
\(\begin{array}{l}\mathbf{\left (a + b \right )^{3}} = a^{3} + b^{3} + 3ab\left (a + b \right )\end{array} \) |
\(\begin{array}{l}\mathbf{\left (a – b \right )^{3} }= a^{3} – b^{3} – 3ab\left (a – b \right )\end{array} \) |
\(\begin{array}{l}\mathbf{(x + y + z)^{2}}= x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2xz\end{array} \) |
\(\begin{array}{l}\mathbf{(x + y – z)^{2}}= x^{2} + y^{2} + z^{2} + 2xy – 2yz – 2xz\end{array} \) |
\(\begin{array}{l}\mathbf{(x – y + z)^{2} } = x^{2} + y^{2} + z^{2} – 2xy – 2yz + 2xz\end{array} \) |
\(\begin{array}{l}\mathbf{(x – y – z)^{2}}= x^{2} + y^{2} + z^{2} – 2xy + 2yz – 2xz\end{array} \) |
\(\begin{array}{l}\mathbf{x^{3} + y^{3} + z^{3} – 3xyz }= (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz -xz)\end{array} \) |
\(\begin{array}{l}\mathbf{x^{2} + y^{2}} = \frac{1}{2} \left [(x + y)^{2} + (x – y)^{2} \right ]\end{array} \) |
\(\begin{array}{l}\mathbf{(x + a) (x + b) (x + c) }= x^{3} + (a + b +c)x^{2} + (ab + bc + ca)x + abc\end{array} \) |
\(\begin{array}{l}\mathbf{x^{3} + y^{3}}= (x + y) (x^{2} – xy + y^{2})\end{array} \) |
\(\begin{array}{l}\mathbf{x^{3} – y^{3}}= (x – y) (x^{2} + xy + y^{2})\end{array} \) |
\(\begin{array}{l}\mathbf{x^{2} + y^{2} + z^{2} -xy – yz – zx }= \frac{1}{2} [(x-y)^{2} + (y-z)^{2} + (z-x)^{2}]\end{array} \) |
| Linear Equation in Two Variables |
\(\begin{array}{l}\mathbf{a_{1}x + b_{1}y + c_{1} }= 0\end{array} \) |
\(\begin{array}{l}\mathbf{a_{2}x+ b_{2}y + c_{2}} = 0 \end{array} \) |
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Excellent work !!!!!!!!!!!!!!
Keep it up
thanks for the formulas
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Thanks for your great work
very helpful thanks a lot
Thanks 😊 for help in study