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Solving Linear Equations (Review)

A linear equation is an equation that can be expressed in the form:
ax+b=0\boxed{ax+b=0}
where a, bR a,~b\in\mathbb{R}~ and 'xx' is the variable.

Steps to Solving a Linear Equation

  1. If there are any fractions in the equation, eliminate them by multiply the equation by the lowest common denominator (LCD).
  2. Eliminate parentheses & combine like terms
  3. Gather all terms with the variable on one side of the equals sign and all other terms on the other side
  4. Divide by the leading coefficient if it is not 1
  5. Verify answer


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Example

Solve x23+1=2x7\displaystyle \frac{x-2}{3}+1=\frac{2x}{7}:

(x23+1=2x7)×217(x2)+21=6x7x14+21=6x(7x6x)=(1421)x=7\displaystyle \begin{array}{r l} \Bigg(\displaystyle \frac{x-2}{3}+1&=\displaystyle \frac{2x}{7}\Bigg)\times21\\\\ 7(x-2)+21&=6x\\\\ 7x-14+21&=6x\\\\ (7x-6x)&=(14-21)\\\\ x&=-7 \end{array}
Verify:
x23+1=2x7723+1=2(7)793+1=23+1=22=2\displaystyle \begin{array}{c l c} \displaystyle \frac{x-2}{3}+1&=\displaystyle \frac{2x}{7}\\\\ \displaystyle \frac{-7-2}{3}+1&=\displaystyle \frac{2(-7)}{7}\\\\ -\displaystyle \frac{9}{3}+1&=-2\\\\ -3+1&=-2\\\\ -2&=2&\checkmark\end{array}

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Example: Solving Linear Equations

Find 'xx':
2x+2=3x1\displaystyle\frac{2}{x+2}=\frac{3}{x-1}

(2x+2=3x1)×(x+2)(x1)2(x1)=3(x+2)2x2=3x+6x=8\begin{array}{rcl} \Bigg(\displaystyle\frac{2}{x+2}&=&\displaystyle\frac{3}{x-1}\Bigg)\times(x+2)(x-1)\\\\ 2(x-1)&=&3(x+2)\\\\ 2x-2&=&3x+6\\\\x&=&-8 \end{array}
Verfiy:

2x+2=3x128+2=38113=13\begin{array}{rclc} \displaystyle\frac{2}{x+2}&=&\displaystyle\frac{3}{x-1}\\\\ \displaystyle\frac{2}{-8+2}&=&\displaystyle\frac{3}{-8-1}\\\\ \displaystyle-\frac{1}{3}&=&\displaystyle-\frac{1}{3}&\checkmark \end{array}

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Solving Quadratic Equations

A quadratic equation is an equation that can be expressed in the form:
ax2+bx+c=0ax^2+bx+c=0
where a,b,c,dRa, b, c, d\in\mathbb{R} and 'xx' is the variable.

Algebraic Methods to Solve a Quadratic Equation

  • Factor
  • Complete the Square
  • Quadratic Formula

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Example
Solve 2x2+x3=02x^2+x-3=0 by factoring, completing the square, & using the quadratic formula

Factoring:
2x2+x3=0(2x22x)+(3x3)=02x(x1)+3(x1)=0(x1)(2x+3)=0 x=1, 32\begin{array}{r c l} 2x^2+x-3&=&0\\\\ (2x^2-2x)+(3x-3)&=&0\\\\ 2x(x-1)+3(x-1)&=&0\\\\ (x-1)(2x+3)&=&0\\\\ \therefore~x=1,~-\frac{3}{2} \end{array}

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Completing the Square:
2x2+x3=0(2x2+x)3=02(x2+12x+116)318=02(x+14)2=258(x+14)2=2516x+14=±54x=1,32\begin{array}{rcl} 2x^2+x-3&=&0\\\\ (2x^2+x)-3&=&0\\\\ 2\big(x^2+\frac{1}{2}x+\frac{1}{16}\big)-3-\frac{1}{8}&=&0\\\\2\big(x+\frac{1}{4}\big)^2&=&\frac{25}{8}\\\\ \big(x+\frac{1}{4}\big)^2&=&\frac{25}{16}\\\\ x+\frac{1}{4}&=&\pm\frac{5}{4}\\\\x&=&1,-\frac{3}{2} \end{array}

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Quadratic Formula:
x=b±b24ac2ax=1±124(2)(3)2(2)x=1±254x=1±54x=1, 32\begin{array}{rcl} x&=&\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x&=&\displaystyle\frac{-1\pm\sqrt{1^2-4(2)(-3)}}{2(2)}\\\\ x&=&\displaystyle\frac{-1\pm\sqrt{25}}{4}\\\\ x&=&\displaystyle\frac{-1\pm5}{4}\\\\ x&=&1,~-\frac{3}{2} \end{array}

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Example: Solving Quadratic Equations (Review)

Solve for 'xx': 2(x+1)2=8-2(x+1)^2=-8

2(x+1)2=8(x+1)2=4(x+1)2=4x+1=±2x=1, 3\begin{array}{rcl} -2(x+1)^2&=&-8\\\\(x+1)^2&=&4\\\\ \sqrt{(x+1)^2}&=&\sqrt{4}\\\\ x+1&=&\pm2\\\\x&=&1,~-3 \end{array}


Solve for 'x':
x2x110=0x^2-x-110=0
(Select all that apply)


Find the roots of 3x24x11=03x^2-4x-11=0.

Solve:
3x+5=(x1)253x+5=(x-1)^2-5

Extra Practice