Difference between revisions of "Substitution"

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Substitution is a relatively universal method to solve simultaneous equations.  It is generally introduced in a first year high school algebra class.  A solution generally exists when the number of equations is exactly equal to the number of unknowns. The method of solving by substitution includes:
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'''Substitution''' is when one replaces all instances of a variable (or expression) with another equivalent variable (or expression).
  
1. Isolation of a variable
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==Uses==
2. Substitution of variable into another equation to reduce the number of variables by one
 
3. Repeat until there is a single equation in one variable, which can be solved by means of other methods. 
 
  
Example:
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===System of Equations===
  
Solve <math>x+y=-1, 3x-y=5</math> for <math>(x,y)</math>. 
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{{main|System of equations}}
  
Start with <math>x+y=-1</math>.   
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Substitution is a relatively universal method to solve simultaneous equationsIt is generally introduced in a first year high school algebra class.  A solution generally exists when the number of equations is exactly equal to the number of unknowns.  The method of solving by substitution includes:
  
<math>x+y=-1</math>          Subtract <math>x</math> from both sides.
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1. Isolation of a variable
  
<math>y=-x-1</math>          <math>y</math> is now isolated.
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2. Substitution of variable into another equation to reduce the number of variables by one
  
Substitute <math>(-x-1)</math> for the y in <math>3x-y=5.</math>
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3. Repeat until there is a single equation in one variable, which can be solved by means of other methods.
  
<math>3x-(-x-1)=5 </math>   Distribute the negative sign.   
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For example, consider the below system.
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<cmath>\left\{\begin{array}{l}x+y=-1\\3x-y=5\end{array}\right.</cmath>
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An example of solving the system by substitution is when we start by isolating <math>y</math> in the top equation to get <math>y = -x - 1</math>.  Then, we can replace all instances of <math>y</math> with <math>(-x-1)</math> in the second equation.  Doing so results in an equation with one variable, and solving it results in
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<cmath>\begin{align*}
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3x-(-x-1) &= 5 \\
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3x+x+1 &= 5 \\
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4x+1 &= 5 \\
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4x &= 4 \\
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x &= 1.
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\end{align*}</cmath>
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After solving for <math>x</math>, we can "plug in" the value <math>1</math> for <math>x</math> to get <math>y = -(1)-1 = -2</math>, so the solution to the system is <math>(1,-2)</math>As usual, we can check by substituting <math>1</math> for <math>x</math> and <math>-2</math> for <math>y</math>.
  
<math>3x+x+1=5</math>        Combine like terms.
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This same method is used for simultaneous equations with more than two equations.
  
<math>4x+1=5</math>          Subtract 1 from both sides.
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===Observing Common Parts===
  
<math>4x=4</math>            Divide both sides by four.
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Substitution can also be used when an expression has multiple common parts.
  
<math>x=1</math>            
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For instance, consider the equation <math>4^x - 6 \cdot 2^x + 8 = 0</math>.  Note that with exponent properties, we can rewrite the equation as <math>(2^x)^2 - 6 \cdot 2^x + 8 = 0</math>.  Because there are multiple instances of <math>2^x</math> in the equation, we can let <math>y = 2^x</math> and substitute <math>2^x</math> for <math>y</math> to make the equation easier to solve.
  
x is now solved for, so substitute x into one of the original equations.   
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Doing so results in
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<cmath>\begin{align*}
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y^2 - 6y + 8 &= 0 \\
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(y-4)(y-2) &= 0 \\
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y &= 2, 4.
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\end{align*}</cmath>
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Finally, we can substitute <math>y</math> for <math>2^x</math> to get <math>2^x = 2</math> or <math>2^x = 4</math>, resulting in <math>x = 1</math> or <math>x = 2</math>As usual, we can check by substituting <math>1</math> and <math>2</math> for <math>x</math> separately.
  
1+y=-1          Subtract 1 from both sides.
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==Problems==
  
y=-2
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===Introductory===
  
<math>(x,y)=(1,-2)</math>
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* [[1961 AHSME Problems/Problem 5]]
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* [[1969 AHSME Problems/Problem 17]]
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* [[2007 iTest Problems/Problem 2]]
  
You can check this answer by plugging x and y into the original equations. 
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===Intermediate===
  
This same method is used for simultaneous equations with more than two equations.
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* [[1983 AIME Problems/Problem 3]]
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[[Category:Algebra]]

Latest revision as of 16:53, 10 April 2020

Substitution is when one replaces all instances of a variable (or expression) with another equivalent variable (or expression).

Uses

System of Equations

Main article: System of equations

Substitution is a relatively universal method to solve simultaneous equations. It is generally introduced in a first year high school algebra class. A solution generally exists when the number of equations is exactly equal to the number of unknowns. The method of solving by substitution includes:

1. Isolation of a variable

2. Substitution of variable into another equation to reduce the number of variables by one

3. Repeat until there is a single equation in one variable, which can be solved by means of other methods.

For example, consider the below system. \[\left\{\begin{array}{l}x+y=-1\\3x-y=5\end{array}\right.\] An example of solving the system by substitution is when we start by isolating $y$ in the top equation to get $y = -x - 1$. Then, we can replace all instances of $y$ with $(-x-1)$ in the second equation. Doing so results in an equation with one variable, and solving it results in \begin{align*} 3x-(-x-1) &= 5 \\ 3x+x+1 &= 5 \\ 4x+1 &= 5 \\ 4x &= 4 \\ x &= 1. \end{align*} After solving for $x$, we can "plug in" the value $1$ for $x$ to get $y = -(1)-1 = -2$, so the solution to the system is $(1,-2)$. As usual, we can check by substituting $1$ for $x$ and $-2$ for $y$.

This same method is used for simultaneous equations with more than two equations.

Observing Common Parts

Substitution can also be used when an expression has multiple common parts.

For instance, consider the equation $4^x - 6 \cdot 2^x + 8 = 0$. Note that with exponent properties, we can rewrite the equation as $(2^x)^2 - 6 \cdot 2^x + 8 = 0$. Because there are multiple instances of $2^x$ in the equation, we can let $y = 2^x$ and substitute $2^x$ for $y$ to make the equation easier to solve.

Doing so results in \begin{align*} y^2 - 6y + 8 &= 0 \\ (y-4)(y-2) &= 0 \\ y &= 2, 4. \end{align*} Finally, we can substitute $y$ for $2^x$ to get $2^x = 2$ or $2^x = 4$, resulting in $x = 1$ or $x = 2$. As usual, we can check by substituting $1$ and $2$ for $x$ separately.

Problems

Introductory

Intermediate

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