Category Archives: Equations

Solving an Exponential Equation: Relating the Bases

Example:  Solve the exponential equation.

4^{x-9}=1/1024

Solution:

The exponential equation
4^{x-9}=1/1024
Try to write both sides of the equation with the same base.  Try 4 since there is a base of 4 on the left 4^{x-9}=1/4^5
Using a property of negative exponents move the base to the numerator  4^{x-9}=4^{-5}
Now that that the bases are the same the exponents must be equal  4^{x-9}=4^{-5}

x-9=-5

Solve for x
 x-9=-5

x-9+9=-5+9

x=4

The solution the the exponential equation is 4.

 

Rational Equation (no solution)

Example: Solve the rational equation.

{x+4}/{x-3}+1=7/{x-3}

Solution:

{x+4}/{x-3}+1=7/{x-3}

Since we are solving a rational equation we need to first find the restrictions (the values of x that cause the expression to be undefined).

To find the restrictions create an equation by setting each denominator equal to zero and solving.

x-3=0

x-3+3=0+3

x=3

Having x=3 causes a zero in the denominator and the overall expression undefined.  That makes 3 a restricted value .

With the restriction in mind we will solve the equation.

 

The original equation
{x+4}/{x-3}+1=7/{x-3}
Multiply each side of the equation by the least common multiple of the denominators.  For this equation the least common multiple is x-3
(x-3)({x+4}/{x-3}+1)=(x-3)7/{x-3}
Distribute the least common multiple to each term.
(x-3){x+4}/{x-3}+1(x-3)=(x-3)7/{x-3}
Simplify by canceling the common factors.  This should clear any denominators.
x+4+1(x-3)=7
Use the distributive property to simplify.
x+4+x-3=7
Simplify each side of the equation by combining like terms.
2x+1=7
Solve for x by getting x by itself on one side.  Start by subtracting 1 on both sides.
2x+1-1=7-1
2x=6
 Solve for x by getting x by itself on one side.  Next divide both sides by 2.
{2x}/2=6/2
x=3
Compare your solution to the restricted value.
Since the solution is the same as the restricted value we must exclude it as a solution.  Since all of the solutions have been excluded, there is no solution to the rational equation.

Video Example:

Higher Order Equation that reduces to a linear equation

Example: Solve the equation.

(x+5)^3-9=x(x+7)(x+8)-6

Solution:

The original equation
(x+5)^3-9=x(x+7)(x+8)-6
Simplify both sides of the equation.  On the left hand side, rewrite the exponent.  On the right hand side, begin to simplify the multiplication.
(x+5)(x+5)(x+5)-9=x(x^2+8x+7x+56)-6
Simplify both sides of the equation.  On the left hand side, begin multiplying.  On the right hand side, combine like terms.
(x+5)(x^2+5x+5x+25)-9=x(x^2+15x+56)-6
Simplify both sides of the equation.  On the left hand side, combine like terms.  On the right hand side use the distributive property.
(x+5)(x^2+10x+25)-9=x^3+15x^2+56x-6
Simplify both sides of the equation.  On the left hand side, continue multiplying.  The right hand side is in simplest form.
x(x^2+10x+25)+5(x^2+10x+25)-9=x^3+15x^2+56x-6
x^3+10x^2+25x+5x^2+50x+125-9=x^3+15x^2+56x-6
Simplify both sides of the equation.  On the left hand side, combine like terms.  The right hand side is in simplest form.
x^3+15x^2+75x+116=x^3+15x^2+56x-6
Now that each side is in simplest form we want the terms with x on one side and the constant terms on the the other side.  Subtract x^3 from each side.  It cancels from each side.
x^3-x^3+15x^2+75x+116=x^3-x^3+15x^2+56x-6
15x^2+75x+116=15x^2+56x-6
Subtract 15x^2 from each side.  It cancels from each side.
15x^2-15x^2+75x+116=15x^2-15x^2+56x-6
75x+116=56x-6
Subtract 56x from each side and simplify. 
75x-56x+116=56x-56x-6
19x+116=-6
Subtract 116 from each side and simplify.
19x+116-116=-6-116
19x=-122
Get x by it self by dividing by 19 on both sides and simplify.
{19x}/19={-122}/19
x={-122}/19

Finding the equation of a line perpendicular to another line

Example: Find the equation of a line perpendicular to another line and passing through a specific point. (The other line in slope intercept form)

Example: Find the equation of a line perpendicular to another line and passing through a specific point. (The other line in slope standard form)

Example: Find the equation of a line perpendicular to the x-axis.

Example: Find the equation of a line perpendicular to the x-axis and perpendicular to the y-axis.

 

Finding the Equation of a Line parallel to another line

Example: Find the equation of the line parallel to another line and passing through a specific point. (parallel equation in slope intercept form)

Example: Find the equation of the line parallel to another line and passing though a specific point. (parallel line in standard form)

Example: Find the equation of the line parallel to the x-axis or y-axis and passing through a specific point.

Example: What is an equation parallel to the y-axis?

Example: What is an equation parallel to the x-axis?

 

Finding the Equation of a line given a fractional slope and a point

Example:  Find the equation of a line in slope intercept form given the slope of the line is -{2/3} and the line passes through the point (-4,7)

Solution:

Use the point-slope formula of the line to start building the line.  m represents the slope of the line and (x_1,y_1) is a point on the line.

Point-slope formula: y-y_1 = m(x-x_1)

m=-{2/3} and  (-4,7)

Substitute the values into the formula.

 y-7 = -{2/3}(x-(-4))

Since the instructions ask to write the equation in slope intercept form (y=mx+b) we will simplify and write the equation with y by itself on one side.  I will also use the clearing fractions method to avoid having to add fractions.

 y-7 = -{2/3}(x-(-4))

3(y-7) = 3[-{2/3}(x+4)] (Multiply by LCM)

3(y-7) = -2(x+4) (Cancel Denominator)

3y-21 = -2x-8

3y-21+21 = -2x-8+21

3y = -2x+13

{3y}/3 = {-2x}/3+{13}/3

y = -{2/3}x+{13}/3

The equation of a line in slope intercept form with a slope of -{2/3} and  passing through the point (-4,7) is   y = -{2/3}x+{13}/3

Finding the Equation of a Line given two points on the line

Example:  Find the equation of a line in slope intercept form given the line passes through the two points (5,-3) and (6,-1).

Solution:

First find the slope of the line.

Choose one of the points to be   ( x_1, y_1) and choose the other point to be   ( x_2, y_2).

I will choose   ( 5, -3)  to be   ( x_1, y_1)  and choose   ( 6, -1) to be   ( x_2, y_2).

Substitute these values into the slope formula and simplify.

  m= {y_2-y_1} / {x_2-x_1} ={-1-(-3)}/{6-5}={-1+3}/{1} =2/1=2

The slope of the line containing the points   ( 5, -3) and   ( 6, -1)  is m= 2.

Then, use the point-slope formula of the line to start building the line.  m represents the slope of the line and you can use (x_1,y_1) or (x_2,y_2) as the point on the line.

Point-slope formula: y-y_1 = m(x-x_1)

m=2 and  (5,-3)

Substitute the values into the formula.

 y-(-3) = 2(x-5)

Since the instructions ask to write the equation in slope intercept form (y=mx+b) we will simplify and write the equation with y by itself on one side.

 y-(-3) = 2(x-5)

 y+3 = 2x-10

 y+3-3 = 5x-10-3

 y = 5x-13

The equation of a line in slope intercept form passing through the two points (5,-3) and (6,-1) is  y = 5x-13.