Tag Archive: Algebra


Various ways to solve for a linear equation

Filed Under: Uncategorized by oscarvalles — Comments Off
November 27, 2010

Linear equations can be presented to you in various ways and by the same token you can solve a linear equation a number of ways. The first step in solving a linear equation is determining the slope.
Here is a list of ways to arrive at the slope.
 
1.)    Given a graphed line, choose two points on the line. Count the rise on the y-axis over the run on the x-axis. Essentially, the slope, m, = rise/run.
2.)    Starting with the coordinates of two points on line (x1, y1) and (x2, y2) use the formula: m = (y2 – y1) / (x2 – x1)
3.)    Rewrite your equation in the form of y = mx + b. m, the coefficient of x, is your slope.
4.)    A line with an unknown slope that is parallel to a line with slope m1, has the same slope as that line.
5.)    A line with an unknown slope that is perpendicular to a line with slope m1 will have a slope equal to the opposite of the reciprocal of m1, that is -1/ m1.
 
 
Next, look at the data points that you have to work with. For example, if you are told that you have $150 to purchase bolts that cost $2 and washers that cost $3, then let x represent bolts and y represent washers. Therefore, your purchasing power is:
 
$150 = 2x + 3y.   You can find two sets of coordinates by setting y = 0 and solving for x and x = 0 and solve for y, which will give you (75, 0) and (0, 50) respectively. Given these two points, you can solve for the slope using item number two, listed above.   
 
m = (50 – 0) / (0 – 75) = -2/3
 
Once you calculate the slope, you will be ready to solve for your equation. 
To do that, use the following method where cp = coordinate point and m=slope:
 
(y – cpy) / (x –cpx) = m
(y – 50) / (x -0) = -2/3
 
Steps to solve for the equation:
 
1.)    (y – 50) / (x-0) = -2/3
2.)    y – 50 = -2/3x
3.)    y = -2/3x + 50
Voila, your equation!
 
Alternatively, and more even more quickly, you could have taken the original equation:
$150 = 2x + 3y
and taken the following steps to solve it:
 
1.)    3y = -2x + 150 (subtract 2x from both sides of the equation)
2.)    y = -2/3x + 50 (divide out the three on both sides of the equation)
 
That will give you the slope intercept form y = -2/3x + 50
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Examples graphs of equations and inequalities

Filed Under: Uncategorized by oscarvalles — Comments Off
May 6, 2009

Below are some example graphs of common equations and inequalities that you are likely to encounter.  

When looking at these charts, notice how inequalities with the less than or equal to sign or the greater than or equal to sign consist of a solid line.  That means that the points in that equation lie not only in the shaded region of the graph but also the line itself.  While inqualities such as greater than or less than will consist of a dotted line.  This means that the vectors in the inequality will never be on the line but rather above or below the line, that is, in the shaded region.  

One way to utilize these charts is to set up your equation in slope intercept form when both the dependent variable, independent variable and constant are present.  If you don’t have all three values, then look for the chart that best represents your equation.  Write out your equation by trying to keep y on one side of the equation or inequality and x and/or k on the other.  If y is not present, then try getting your equation or inequality where x is on one side and k on the other.  Search for the sample graph below.  If you are tackling a word problem or real life problem, ask yourself if it makes sense for your data points (vectors) to lie in the shaded regions (or solid line when present).

Without wasting more time, the chart examples are presented below.

 

 

 

 

 

 

y = mx + b

 

y = mx + b

 

 

 

 

 

 

y >= mx + b

 

y is greather than or equal to mx + b

 

 

 

 

 

 

y <= mx + b

 

y is less than or equal to mx + b

 

 

 

 

 

 

 

y > mx + b

 

y is greater than mx + b

 

 

 

 

 

 

y < mx + b

 

y is less than mx + b

 

 

 

 

 

 

y = k

 

y = k; where k is a constant number

 

 

 

 

 

 

y >= b

 

y is greater than or equal to b; where b is a constant number

 

 

 

 

 

 

y <= b

 

y is less than or equal to b; where b is a constant

 

 

 

 

 

 

y > b

 

y is greater than b; where b is a constant

 

 

 

 

 

 

y < b

 

y is less than b; where b is a constant

 

 

 

 

 

 

x = k

 

x is equal to k; where k is a constant

 

 

 

 

 

 

x >= k

 

x is greater than or equal to k;; where k is a constant

 

 

 

 

 

 

x <= k

 

x is less than or equal to k, where k is a constant

 

 

 

 

 

 

x > k

 

x is greater than kl; where k is a constant

 

 

 

 

 

 

x < k

 

x is less than k; where k is a constant

 

 

 

 

 

 

That is pretty much it.  If you need a graph to a particular equation or inequality, leave a comment with your equation and I will do my best to reply asap with a solution.

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Diluted Mixtures in Algebraic Terms

Filed Under: Uncategorized by oscarvalles — Comments Off
March 15, 2009

 The following situation requires you to create 30 mL of a solution that contains 10% acid.  You have an abundance of 5% and 20 % acid solutions on hand to create the 30mL 10% solution with.  The question is, how many mL of each solution should be mixed together to form the 30mL 10% solution?

We can express the paragraph above by saying: 30mL of 10% solution equals x mL of 20% solution plus y mL of 5% solution.

Let’s begin by identifying the outcome we seek.  This would be the 30mL of a 10% solution.  We’ll place that on the one side of the equal sign.  Place it on the side you are more comfortable with (left or right).

(30)(.10) =  a mL of 20% solution plus b mL of 5% solution

Next let’s begin writing the part of the equation that is opposite to our 30mL 10% solution (i.e. (30)(.10) ).

We know we cannot exceed 30 gallons of either the 5% or 20% solution in our mixture so we’ll have to incorporate this into our equation.  If you notice above, we denote the milliliters we need from both solutions by a and b.  Since a and b are not like terms, we will need to find a way to convert them to like terms.  Knowing that we need our mixitures to result in exactly 30mL, we can convert a and b to (30mL – x) and (30mL-(30mL-x)) respectively.  The two expression complement themselves inversely and when added equal 30mL.

Our new equation is: (30mL)(.10) = .05(30mL – x) and .20(30mL-(30mL-x)) 

Working it in steps we have the following:

  1. (30)(.10) = .05(30 – x) + .20(30-(30-x)) 
  2. 3 = 1.5 – .05x + 6 – 6 + .20x
  3. 3 = 1.5 + .15x
  4. 1.5 = .15x
  5. 10 = x

Now we see that 10mL is equal to x.  If we plug this in to our equation, for the 5% solution we’ll know that 20mL are supposed to be used: .05(30mL – 10mL) = .05(20mL).  Fill in the x for the 20% solution and you’ll find that 10mL of the 20% solution is needed. .20(30ml – (30 mL – 10 mL)) -> .20(30mL – (20mL)) -> .20(10mL)

You can verify this by writing the working out the equation with x = 10.  The result will be 30mL at 10% solution – or – 3.

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