Watch Video on Coin Word Problems – Algebra Help
Tuesday, March 2nd, 2010
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/coinproblems.php Students learn to solve “value” word problems, such as the following. Martin has a total of 19 nickels and dimes worth $1.65. How many of each type of coin does he have? Note that this problem requires a chart to organize the information. The chart is based on the total value formula, which states that the number of coins times the value of each coin = the total value. The chart is then used to set up the equation.
Duration : 0:4:50
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/graphingsystemsofequations.php Students learn to solve a system of linear equations by graphing. The first step is to graph each of the given equations, then find the point of intersection of the two lines, which is the solution to the system of equations. If the two lines are parallel, then the solution to the system is the null set. If the two given equations represent the same line, then the solution to the system is the equation of that line. Visit http://www.yourteacher.com to browse our huge library of online math videos and try a free online lesson!
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/solvinginequalities.php Students learn that when solving an inequality, such as –3x is less than 12, the goal is the same as when solving an equation: to get the variable by itself on one side. Note that when multiplying or dividing both sides of an inequality by a negative number, the direction of the inequality sign must be switched. For example, to solve –3x is less than 12, divide both sides by –3, to get x is greater than -4. And when graphing an inequality on a number line, less than or greater than means an open dot, and less than or equal to or greater than or equal to means a closed dot.
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/completingthesquare.php Students learn to solve quadratic equations by completing the square. For example, to solve the equation s^2 – 6s + 5 = 0, the first step is put the constant term on the opposite side of the equation as the terms that contain the variables, by subtracting 5 from both sides, to get s^2 – 6s ___ = – 5 ___. Next, the number that goes in each space comes from half the coefficient of the middle term squared, which in this case is half of -6, or -3, squared, which is +9. So a +9 goes in each space, to get s^2 – 6s + 9 = – 5 + 9. The trinomial on the left side of the equation then factors as (s – 3)(s – 3), or (s – 3)^2, and the right side of the equation simplifies to 4, so the problem now reads (s – 3)^2 = 4. Finally, take the square root of both sides to get s – 3 = plus or minus 4, so s – 3 = 4, or s – 3 = -4, and solving each equation from here, s = 7 or -1.
Access full lesson containing this video at: http://www.yourteacher.com/algebra2/thequadraticformula.php Students learn to solve quadratic equations in the form ax^2 + bx + c = 0 using the quadratic formula, which states that x = -b plus or minus the square root of b^2 — 4ac over 2a. For example, to solve the equation x^2 — 3x — 8 = 0, since a = 1, b = -3, and c = -8, the quadratic formula states that x = 3 plus or minus the square root of (-3)^2 — 4(1)(-8) over 2(1). Simplifying on the right side of the equation, x = 3 plus or minus the square root of 9 + 32 over 2, or x = 3 plus or minus the square root of 41 over 2, which is the final answer. Note that students also learn to derive the quadratic formula using completing the square in the first example problem in this lesson.