Determine whether each ordered pair is a solution of the system of linear equations. See Example

Determine whether each ordered pair is a solution of the...

Key Concepts

Substitution and Verification

The substitution method involves plugging a potential solution or known value into the equations to verify if the equality holds. This process is used to check whether an ordered pair truly satisfies each equation of the system. It's a straightforward yet essential technique for confirming the validity of a solution to a system of equations.

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. The solution to the system is a set of values for the variables that satisfy all equations simultaneously. This concept is central to algebra and appears in various applications, such as modeling real-world problems where multiple conditions must be met concurrently.

Ordered Pair

An ordered pair is a pair of numbers used to represent the values of variables, typically in the form (x, y) for two-dimensional systems. These pairs indicate specific points on the coordinate plane and serve as potential solutions to equations or systems of equations. Understanding ordered pairs is fundamental when evaluating whether a given pair satisfies a particular equation.

Transcript

00:02 All right, so for number seven, we're checking to see if the points given in part a and part b are solutions to the system of linear equations, the same way that we've done in the first six problems.

00:15 But this time we have some fractions, which makes things a little bit more complicated.

00:21 But the rules are the same.

00:23 If we plug in x and y and the equation is true, then therefore it becomes a solution to that system of equations.

00:30 As long as it's true for both of those equations.

00:34 So starting with part a, my x value is negative 2 and my y value is 0.

00:38 I'm going to plug those into the first equation and see if it's true.

00:43 So negative 2 is equal to x, which is negative 2, minus 7 times my y of 0.

00:50 Now this is, so on the right side we have to first do negative 7 times 0, which is 0.

00:57 So all i'm left with is a negative 2.

00:59 So negative 2 is equal to negative 2.

01:01 That's a true statement, which means i need to check that point for my second line and make sure it works for that one, and then it would be a solution.

01:12 So my second line is 6x minus y equals 13.

01:15 I'm going to plug in my x and my y, so i'm going to do 6 times negative 2 minus 0 and see if that's equal to 13.

01:25 6 times negative 2 is negative 12, so negative 12 is not equal to 13.

01:32 So since that is incorrect, that means that negative 2 -0 is not a solution.

01:47 All right, so let's check another one for part b.

01:56 In this case, i'm using the fraction 1 -half as my x value and the fraction 514th as my y value.

02:06 And i'm plugging into each equation again.

02:09 So my first equation is negative 2 equals x minus, 7y.

02:16 So that means that x is one half minus seven times my y, which is 5 over 14.

02:26 Now i'm going to convert this 7 to 7 over 1 so i can multiply those two fractions.

02:31 So this becomes one half and then let's look at 7 over 1 times 5 over 14.

02:41 So i can simplify those two fractions since we're multiplying them together, the 7 can just become a 1, and the 14 a 2, because 7 over 14 would be 1 half.

02:56 So then it makes my multiplication a lot easier.

02:59 So i have, basically i have negative 1 times 5, so that's going to be my minus.

03:04 1 times 5 is 5 over 1 times 2, which is 2.

03:11 So that's good because now my denominators match and i can go ahead and simplify.

03:18 The right hand side i have one half minus five halves.

03:21 So one minus five is four...

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