Simultaneous Equations - Tons of Examples!

This paragraph introduces the topic of solving simultaneous equations using various methods such as substitution, elimination, graphing, and handling systems with three variables. It begins with a two-variable system and the elimination method, using the example of two equations: 2x + y = 4 and 2x - y = 0. The process involves adding the equations to eliminate the variable y and solve for x, followed by substituting x back into one of the original equations to solve for y, resulting in the solution (1,2). The paragraph continues with another example using the elimination method with different coefficients and suggests the viewer to try an example themselves, pausing the video to work it out.

The second paragraph delves into a more complex example of the elimination method, where the equations are 3x + 2y = 12 and 4x - y = 5. The speaker explains that adding these equations directly won't eliminate any variables, so they suggest multiplying the second equation by 2 to make the y terms cancel out. After rewriting and adding the modified equations, the result is 11x = 22, leading to x = 2. Subsequently, the value of x is substituted back into the first equation to solve for y, yielding y = 3. The final answer is given as the ordered pair (2,3). The paragraph also introduces a problem involving fractions and suggests clearing the fractions by finding a common denominator.

In this paragraph, the focus is on solving a system of equations with fractions: (2/3)x + (1/4)y = 9 and (5/9)x - (5/6)y = -5. The speaker recommends clearing the fractions by finding the least common multiple (LCM) of the denominators, which in this case is 12 for the first equation and 18 for the second. The equations are then multiplied by these LCMs to eliminate the fractions, resulting in 8x + 3y = 108 and -10x + 15y = 90. The first equation is further multiplied by 5 to align the y terms for cancellation. After adding the equations, x is solved to be 9. Substituting x back into the first equation yields y = 12, and the solution is the point (9,12).

The substitution method is introduced for solving systems of equations where one variable is expressed in terms of the other. Two examples are given: y = 2x - 5 and y = 5x - 11, and y = (1/4)x + 2 and y = (3/5)x - 5. In both cases, the equations are set equal to each other, and the resulting equations are solved for x. Once x is found, it is substituted back into one of the original equations to solve for y. The solutions are (2, -1) and (20, 7), respectively. The speaker also demonstrates how to clear fractions by finding the LCM and multiplying through to eliminate them.

This paragraph presents a scenario where the substitution method is applied to equations with decimal coefficients: y = 0.1x + 0.2 and 0.3x + 3y = 5.4. The speaker suggests eliminating decimals by multiplying everything by 10, transforming the equations into 3x + 30y = 54 and 0.6x + 3y = 6. After combining like terms, x is solved to be 8. Substituting x back into the first equation gives y = 1, and the solution is the point (8,1).

The graphing method for solving systems of equations is introduced. Two examples are given: y = -2x + 7 and y = 3x - 3, and y = (1/3)x - 2 and y = -(4/3)x + 3. For each pair of equations, the speaker explains how to graph each line by determining the slope and y-intercept, and then finding additional points by using the slope-intercept method. The solutions are found at the points of intersection of the two lines, which are (2,3) and (3,-1), respectively.

The paragraph discusses solving a system of three equations with three variables: x + y + z = 6, x + y - z = 0, and x - y + 2z = 5. The speaker demonstrates how to eliminate variables by adding pairs of equations to find relationships between x, y, and z. After simplifying, two new equations with two variables each are obtained: 2x + 2y = 6 and 3x + y = 5. The elimination method is then used again to solve for x and y, followed by substituting these values back into one of the original equations to solve for z. The solution is the point (1,2,3).

This paragraph presents additional examples of systems with three variables: 2x + y + 3z = 19, 3x - 2y + 4z = 16, and 5x - y + 2z = 15. The speaker guides the viewer through the process of using the elimination method to cancel variables and solve for x, y, and z. The solutions are found to be x = 2, y = 3, and z = 4, and these values are checked in all three original equations to confirm the solution.

The paragraph applies the concepts of simultaneous equations to word problems. The first problem involves finding two numbers given their sum and difference, resulting in the equations x + y = 26 and x - y = 6. The second problem is about Sally's savings accounts, where she earns a total of $2400 in interest from two accounts with different interest rates, leading to the equations 0.12x + 0.16y = 2400 and x + y = 17000. The third problem involves calculating the number of nickels, dimes, and quarters in a jar given the total number of coins and their total value, resulting in the equations n + d + q = 38, 5n + 10d + 25q = 570, and d = n + 2. The solutions are found by solving these systems of equations.

The final paragraph concludes the video with a brief sign off, thanking viewers for watching and wishing them a great day.