How do I solve an equation with constants in the denominators?
ok, for the 1st problem: so what you want to do first, is make a common denominator for the two fractions. do this by multiplying the left by 2. 2x/4=3x/4+5. move the 3x/4 to the left, and subract. 2x/4-3x/4=5 -x/4=5 cross multiply -x=20 x=-20. to check if this is right, plug it back into the equation. it works! try the 2nd prob, and email me if u cant get the rest! tUNZ
How do I solve an equation with a variable in the numerator?
The way I read the question, your problem seems to be that not everything has the same denominator. Just extend the fractions so that the denominator of all terms are identical.20 - P/2 = 2P<-> 40/2 -P/2 =4P/2Since the denominator is not 0, you can multiply the equation with it and get40 -P =4PNow there are no more fractions in the equation and you can solve it just like you have learned.40=5P8=P
Are any of these linear equations?
yes, no, maybe, so,
How do you write linear equations?
With a pencil…Seriously though, an equation is linear if it contains no variables of order 2 or higher (squared, cubed…), no variables in the denominator, and no transcendental functions (sin x, ln x, e^x…). It is sometimes referred to as a “first-order” equation (cue Star Wars music…).These are seen quite frequently in everyday life. Imagine the following scenario:Joe is hired to sell cars. He gets a $100 bonus when he’s hired, and makes $50 for each car he sells. How much money does Joe have?It depends on how many cars he sells. How much money he has is a function of how many cars he sells. Let $ be money made and “n” be the number of cars sold. The money in his pocket at any given time is defined by $ = 50n + 100. This is a linear equation in y = mx + b form where 50 is the rate of change ($50 for each car) and 100 is his starting point.
Literal equation - isolating a variable?
I have serious issues with literal equations. I think my problems are in where to add and subtract and when to multiply and divide (to isolate variables and clear denominators). Here is my question: 1/f = 1/f1 + 1/f2 (isolate f1) I have gotten this far, and am not sure if it is even correct: 1/f = 1/f1 + 1/f2 LCD: f f1 f2 after clearing the denominators, I get: f1f2 = ff2 + ff1 I really don't know where to go from here, there are so many fs that are attached to different fs. Can anyone tell me if I am going in the right direction and if not, how to solve this? Thank you!
Linear equations?
Yes, graphing a linear equation on a Cartesian plane results in a line. And yes, linear equations have only powers of one on the exponents. So, "y = 3x" IS a linear equation, but "y = 3x^2" is NOT. But also, linear equations don't have two variables mutiplied together. So, "9x + 3y = 12" IS a linear equation, but "9xy = 12" is NOT. And, linear equations don't have variables in denominators. So, "x = 3" IS a linear equation, but "(1/x) - y = 3" is NOT.
What are linear equations?
A linear equation looks like any other equation. It is made up of two expressions set equal to each other. A linear equation is special because: 1) It has one or two variables. 2) No variable in a linear equation is raised to a power greater than 1 or used as the denominator of a fraction. 3)When you find pairs of values that make the linear equation true and plot those pairs on a coordinate grid, all of the points for any one equation lie on the same line. Linear equations graph as straight lines. A linear equation in two variables describes a relationship in which the value of one of the variables depends on the value of the other variable. In a linear equation in x and y, x is called x is the independent variable and y depends on it. We call y the dependent variable. If the variables have other names, yet do have a dependent relationship, the independent variable is plotted along the horizontal axis. Most linear equations are functions (that is, for every value of x, there is only one corresponding value of y). When you assign a value to the independent variable, x, you can compute the value of the dependent variable, y. You can then plot the points named by each (x,y) pair on a coordinate grid. The real importance of emphasizing graphing linear equations with your students, is that they should already know that any two points determine a line, so finding many pairs of values that satisfy a linear equation is easy: Find two pairs of values and draw a line through the points they describe. All other points on the line will provide values for x and y that satisfy the equation. Describing Linear Relationships The graphs of linear equations are always lines. One important thing to remember about those lines is: Not every point on the line that the equation describes will necessarily be a solution to the problem that the equation describes. Examples of Linear Relationships distance = rate x time In this equation, for any given steady rate, the relationship between distance and time will be linear. However, distance is usually expressed as a positive number, so most graphs of this relationship will only show points in the first quadrant. Notice that the direction of the line in the graph below is from bottom left to top right. Lines that tend in this direction have positive slope. A positive slope indicates that the values on both axes are increasing from left to right.
What is a linear equation
y=mx+q or ax+by+c=0
Is 3/x+4/y=2 a linear equation?
I understand that the standard equation is Ax+By=C but I'm not sure if you can use the variables x and y as denominators in fractions. Please help as soon as possible. Thank you :)