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Adding and Subtracting Polynomials | MathHelp.com
By MathHelp.com
In this example, notice that each of our variables, x, y, and z, appears in all three equations. To solve this system, we use the addition method. In other words, let’s start with our first two equations, x + y + z = 4, and x – y + z = 2. Notice that if we add these equations together, the +y and –y will cancel out, and we have 2x + 2z = 6. So, in our new equation, 2x + 2z = 6, we’ve eliminated the variable y. Unfortunately, we still haven’t solved for any of our variables. However, if we can create another equation with just x and z in it, then we’ll have a system of equations in two variables, which we can use to solve for x and z. To create another equation with just x and z in it, we need to eliminate y. We can’t add the first and second equations together, because we’ve already done that. However, notice that if we add the first and third equations together, the first equation has a +y and the third equation has a –y, so we’ll be able to eliminate the y. So we have our first equation, x + y + z = 4, and our third equation, x – y – z = 0, and adding them together, notice that the +y – y cancels out, and, as a bonus, the +z – z also cancels out, so we have 2x = 4, and dividing both sides by 2, x = 2. Now, since we know that x = 2, notice that if we plug a 2 in for x in the equation that we created earlier, we can solve for z. And we have 2(2) + 2z = 6, or 4 + 2z = 6, and subtracting 4 from both sides, we have 2z = 2, and dividing both sides by 2, z = 1. So x = 2 and z = 1, and to find the value of y, we simply plug our values of x and z into any of the equations in the original system. Let’s use the first equation, x + y + z = 4. Since x = 2 and z = 1, we plug a 2 in for x and a 1 in for z, and we have 2 + y + 1 = 4, or 3 + y = 4, and subtracting 3 from both sides, y = 1. So x = 2, y = 1, and z = 1, and finally, we write our answer as the ordered triple, x, y, z, or (2, 1, 1).
Systems of Three Equations | MathHelp.com
By MathHelp.com
Here we’re asked to graph the following function and use the horizontal line test to determine if it has an inverse. And if so, find the inverse function and graph it. So let’s start by graphing the given function, f(x) = 2x – 4, and remember that f(x) is the same as y, so we can rewrite the function as y = 2x – 4. Now, we simply graph the line y = 2x – 4, which has a y-intercept of -4, and a slope of 2, or 2/1, so we go up 2 and over 1, plot a second point and graph our line, which we’ll call f(x). Next, we’re asked to use the horizontal line test to determine if the function has an inverse. Since there’s no way to draw a horizontal line that intersects more than one point on the function, the function does have an inverse. So we need to find the inverse and graph it. To find the inverse, we switch the x and the y in original function, y = 2x – 4, to get x = 2y – 4. Next, we solve for y, so we add 4 to both sides to get x + 4 = 2y, and divide both sides by 2 to get 1/2x + 2 = y. Next, let’s flip our equation so that y is on the left side, and we have y = 1/2x + 2. Finally, we replace y with the notation that we use for the inverse function of f, as shown here. And remember that we’re asked to graph the inverse as well, so we graph y = ½ x + 2. Our y-intercept is positive 2, and our slope is ½, so we go up one and over 2, plot a second point, graph the line, and label it as the inverse function of f. Notice that the graph of the inverse function is a reflection of the original function in the line y = x.
Absolute value inequalities | Linear equations | Algebra I | Khan Academy
By Khan Academy
This video lecture series on Worked Examples in Algebra from Khan Academy includes Solving Equations, Solving Word Problems, Solving for a variable, Absolute Value and Number Lines, Patterns in Sequences, Functional Relationships, Domain and Range, Rate Problems, Linear Functions, Slope of a Line, X and Y intercepts, Equation of a Line, Parallel Lines, Perpendicular Lines, Solving Inequalities and more...
Add and subtract polynomials with one variable
By Khan Academy
Sal simplifies (x^3 + 3x - 6) + (-2x^2 + x - 2) - (3x - 4).
Add and subtract polynomials with one variable
By Khan Academy
Sal analyzes a polynomial subtraction process to find the step that has an error.
Add and subtract polynomials with one variable
By Khan Academy
Sal simplifies (16x+14) - (3x^2 + x - 9).
Add and subtract polynomials with one variable
By Khan Academy
Sal simplifies (5x^2 + 8x - 3) + (2x^2 - 7x + 13x).
Algebra Basics: What Is Algebra? - Math Antics
By Lumos Learning
This video gives an overview of Algebra and introduces the concepts of unknown values and variables. It also explains that multiplication is implicit in Algebra.
08 - Solving Exponential Equations - Part 1 - Solve for the Exponent
By Math and Science
Quality Math And Science Videos that feature step-by-step example problems!
One-step addition and subtraction equations with fractions and decimals
By Khan Academy
Some quick examples to practice solving a variety of one step equations. All 4 operations (add, subtract, multiple, divide) are paired with variables.
One-step addition and subtraction equations with fractions and decimals
By Khan Academy
Learn how to solve one-step addition and subtraction equations that have fractions and decimals in them.