https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sohanraghavmulinti&feedformat=atomAoPS Wiki - User contributions [en]2021-05-07T19:29:34ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Algebraic_manipulation&diff=137047Algebraic manipulation2020-11-08T22:49:35Z<p>Sohanraghavmulinti: /* Example */</p>
<hr />
<div>'''Algebraic manipulation''' involves doing opposite operations (undoing) to any equation to solve for a certain variable (often by isolation).<br />
<br />
==Properties of Equality==<br />
<br />
Any time we add, subtract, multiply, divide, square, square root, etc. to one side, we must do it to the other side to maintain equality.<br />
<br />
===Example===<br />
<cmath>\begin{align*}<br />
x^2 + 18 &= 43 \\<br />
x^2 &= 25 \\<br />
x &= \boxed{\pm 5}<br />
\end{align*}</cmath><br />
In the example, we first subtract 18 from the left side to isolate the <math>x^2</math>. However, we also have to subtract 18 from the right side to maintain equality. The right hand side becomes <math>43 - 18 = 25</math>.<br />
<br />
We then square root the left side to get the <math>x</math> by itself. However, we also have to square root the right side to maintain equality.<br />
<br />
==Cross Multiplication==<br />
<br />
Cross multiplication is a common method of solving [[proportions]]. Essentially, one is multiplying both sides by the denominators of both sides.<br />
<br />
===Example===<br />
<cmath>\begin{align*}<br />
\frac{x}{3} &= \frac{10}{5} \\<br />
5 \cdot x &= 3 \cdot 10 \\<br />
5x &= 30 \\<br />
x &= 6<br />
\end{align*}</cmath><br />
The above method of solving is an example of cross-multiplication. During the cross-multiplication, we actually multiplied both sides by <math>3 \cdot 5</math>. On the LHS, the <math>3</math> gets divided out (leaving a <math>5</math> for multiplication), and on the RHS, the <math>5</math> gets divided out (leaving a <math>3</math> for multiplication).<br />
<br />
==Potential Extraneous Solutions==<br />
<br />
In some methods of isolation like squaring both sides or multiplying both sides by a non-constant, we could introduce additional "solutions" that aren't really solutions to the original equation. Such solutions are called "extraneous solutions". Thus, when doing such methods, it is a good idea to check the solutions by plugging it back into the original equation.<br />
<br />
Additionally, when dividing both sides by a non-constant, we could eliminate perfectly valid solutions. Thus, when possible, it's often better to [[factoring|factor]] than to divide both sides.<br />
<br />
===Example===<br />
<cmath>\begin{align*}<br />
\frac{x^2 - 2x}{x-5} &= \frac{15}{x-5} \\<br />
x^2 - 2x &= 15 \\<br />
x^2 - 2x - 15 &= 0 \\<br />
(x-5)(x+3) &= 0 \\<br />
x &= 5, -3<br />
\end{align*}</cmath><br />
When solving, we start by multiplying both sides by <math>x-5</math>. Then we used algebraic manipulation to get all terms to one side to form a [[quadratic]]. After [[factoring]], we get <math>x = 5, -3</math>.<br />
<br />
Plugging in <math>x = -3</math> satisfies the original equation because <math>\frac{9+6}{-8} = \frac{15}{-8}</math>. However, plugging in <math>x = 5</math> does not satisfy the original equation because <math>\frac{15}{0}</math> on both sides is undefined. Thus, <math>x = -3</math> is the only valid solution, and <math>x = 5</math> is an extraneous solution.<br />
<br />
The extraneous solution appeared when we multiplied both sides by <math>x - 5</math>. The original equation banned <math>x = 5</math> as a possible solution because it would make the denominator zero. However, after multiplying both sides by <math>x-5</math>, there is no denominator that eliminates <math>x = 5</math> as a possible solution.<br />
<br />
==Problems==<br />
<br />
===Introductory===<br />
* Practice Problems on Khan Academy<br />
** [https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-one-step-add-sub-equationss/e/one_step_equations One Step Addition/Subtraction Linear Equation]<br />
** [https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-one-step-mult-div-equations/e/linear_equations_1 One Step Multiplication/Division Linear Equation]<br />
** [https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-2-step-equations-intro/e/linear_equations_2 Two Step Linear Equation]<br />
* Practice Problems on [https://artofproblemsolving.com/alcumus Alcumus]<br />
** Basic Linear Equations<br />
** Advanced Linear Equations<br />
* [[1950 AHSME Problems/Problem 2]]<br />
* [[1990 AHSME Problems/Problem 1]]<br />
<br />
===Intermediate===<br />
* [[1986 AIME Problems/Problem 1]]<br />
<br />
==See Also==<br />
* [[Distributive property]]<br />
<br />
[[Category:Algebra]]</div>Sohanraghavmulintihttps://artofproblemsolving.com/wiki/index.php?title=Algebraic_manipulation&diff=137046Algebraic manipulation2020-11-08T22:49:05Z<p>Sohanraghavmulinti: /* Potential Extraneous Solutions */</p>
<hr />
<div>'''Algebraic manipulation''' involves doing opposite operations (undoing) to any equation to solve for a certain variable (often by isolation).<br />
<br />
==Properties of Equality==<br />
<br />
Any time we add, subtract, multiply, divide, square, square root, etc. to one side, we must do it to the other side to maintain equality.<br />
<br />
===Example===<br />
<cmath>\begin{align*}<br />
x^2 + 18 &= 43 \\<br />
x^2 &= 25 \\<br />
x &= \boxed{\pm 5}<br />
\end{align*}</cmath><br />
In the example, we first subtract 18 from the left side to isolate the <math>x^2</math>. However, we also have to subtract 18 from the right side to maintain equality. The right hand side becomes <math>43 - 18 = 25</math>.<br />
<br />
We then square root the left side to get the <math>x</math> by itself. However, we also have to square root the right side to maintain equality.<br />
<br />
==Cross Multiplication==<br />
<br />
Cross multiplication is a common method of solving [[proportions]]. Essentially, one is multiplying both sides by the denominators of both sides.<br />
<br />
===Example===<br />
<cmath>\begin{align*}<br />
\frac{x}{3} &= \frac{10}{5} \\<br />
5 \cdot x &= 3 \cdot 10 \\<br />
5x &= 30 \\<br />
x &= 6<br />
\end{align*}</cmath><br />
The above method of solving is an example of cross-multiplication. During the cross-multiplication, we actually multiplied both sides by <math>3 \cdot 5</math>. On the LHS, the <math>3</math> gets divided out (leaving a <math>5</math> for multiplication), and on the RHS, the <math>5</math> gets divided out (leaving a <math>3</math> for multiplication).<br />
<br />
==Potential Extraneous Solutions==<br />
<br />
In some methods of isolation like squaring both sides or multiplying both sides by a non-constant, we could introduce additional "solutions" that aren't really solutions to the original equation. Such solutions are called "extraneous solutions". Thus, when doing such methods, it is a good idea to check the solutions by plugging it back into the original equation.<br />
<br />
Additionally, when dividing both sides by a non-constant, we could eliminate perfectly valid solutions. Thus, when possible, it's often better to [[factoring|factor]] than to divide both sides.<br />
<br />
===Example===<br />
<cmath>\begin{align*}<br />
\frac{x^2 - 2x}{x-5} &= \frac{15}{x-5} \\ i hol they tha you karawe cau i bou oo bea yoo aaaah<br />
x^2 - 2x &= 15 \\<br />
x^2 - 2x - 15 &= 0 \\<br />
(x-5)(x+3) &= 0 \\<br />
x &= 5, -3<br />
\end{align*}</cmath><br />
When solving, we start by multiplying both sides by <math>x-5</math>. Then we used algebraic manipulation to get all terms to one side to form a [[quadratic]]. After [[factoring]], we get <math>x = 5, -3</math>.<br />
<br />
Plugging in <math>x = -3</math> satisfies the original equation because <math>\frac{9+6}{-8} = \frac{15}{-8}</math>. However, plugging in <math>x = 5</math> does not satisfy the original equation because <math>\frac{15}{0}</math> on both sides is undefined. Thus, <math>x = -3</math> is the only valid solution, and <math>x = 5</math> is an extraneous solution.<br />
<br />
The extraneous solution appeared when we multiplied both sides by <math>x - 5</math>. The original equation banned <math>x = 5</math> as a possible solution because it would make the denominator zero. However, after multiplying both sides by <math>x-5</math>, there is no denominator that eliminates <math>x = 5</math> as a possible solution.<br />
<br />
==Problems==<br />
<br />
===Introductory===<br />
* Practice Problems on Khan Academy<br />
** [https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-one-step-add-sub-equationss/e/one_step_equations One Step Addition/Subtraction Linear Equation]<br />
** [https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-one-step-mult-div-equations/e/linear_equations_1 One Step Multiplication/Division Linear Equation]<br />
** [https://www.khanacademy.org/math/pre-algebra/pre-algebra-equations-expressions/pre-algebra-2-step-equations-intro/e/linear_equations_2 Two Step Linear Equation]<br />
* Practice Problems on [https://artofproblemsolving.com/alcumus Alcumus]<br />
** Basic Linear Equations<br />
** Advanced Linear Equations<br />
* [[1950 AHSME Problems/Problem 2]]<br />
* [[1990 AHSME Problems/Problem 1]]<br />
<br />
===Intermediate===<br />
* [[1986 AIME Problems/Problem 1]]<br />
<br />
==See Also==<br />
* [[Distributive property]]<br />
<br />
[[Category:Algebra]]</div>Sohanraghavmulinti