Solve Problems 5-26 using elimination by addition. 2 x+y-z=5 x-2 y-2 z=4 3 x+4 y+

Solve Problems 5-26 using elimination by addition. 2...

Key Concepts

Systems of Linear Equations

A system of linear equations consists of multiple equations with several variables that must all be true simultaneously. The goal is to find values for these variables that satisfy every equation in the system. This concept is foundational in algebra and applied mathematics, as it models situations where multiple linear relationships coexist.

Elimination Method

The elimination method is a technique for solving systems of equations by removing one or more variables to reduce the complexity of the system. This is typically achieved by adding or subtracting equations after appropriately scaling them, so that one of the variables cancels out. This process is repeated until the system is simplified enough to solve for all variables.

Addition Technique in Elimination

The addition technique involves directly adding or subtracting equations once they have been manipulated (usually by multiplying by constants) so that one of the corresponding variable’s coefficients are opposites. This cancels that variable out, making it easier to focus on solving for the remaining variables.

Back-Substitution

Back-substitution is the process of substituting the value of a solved variable back into other equations to determine the values of the remaining variables. After reducing the system and finding a specific variable's value, back-substitution systematically resolves the rest of the variables, ensuring that all solutions satisfy the original system of equations.

Transcript

00:03 Given the system of three equations and three unknowns, we want to solve using the method of elimination.

00:10 So let's start by eliminating the y's by taking our a equation and multiplying it by two.

00:19 So two times the equation a, two times two x is four x, plus two times y is two y, plus two times negative z is negative two z equals two times five.

00:34 Is 10.

00:35 And now we're going to add that to the equation b x minus 2y minus 2z equals 4.

00:46 So we are going to eliminate the ys this way.

00:50 So when we add them together, 4x plus x is 5x plus 0 and negative 2 plus negative 2 is negative 4 z equals 10 plus 4 is 14.

01:04 So 5x minus 4 z is 14.

01:07 And we'll go ahead and give that a name.

01:08 We'll call that equation d.

01:11 Now we can also eliminate the y from the b equation by multiplying the b equation by 2.

01:22 So we're going to take that right here, 2 times b, and we'll have 2x minus 4y minus 4 z equals 2 times 4.

01:35 Is 8.

01:36 Now when we add c and b together, we get 3x plus 2x is 5x, 4y minus 4 way, that's 0, and 3z minus 4z is negative 1z equals 3 plus 8 is 11.

01:54 So 5x minus z is 11, and we will call that equation e.

01:59 Now, when we look at d and e, if we multiply equation d by negative 1, we'll be able to eliminate the x's and solve for z.

02:10 So we'll take negative 1 times d, which will be negative 1 times 5x.

02:17 Negative 1 times negative 4z is plus 4 z, and negative 1 times 14 is negative 14.

02:26 So positive 5 and negative 5x, that goes away.

02:30 Negative z plus 4 z gives us 3 z equals negative 11, excuse me, positive 11 minus 14 is negative 3.

02:40 And solving for that, we find that z equals negative 1.

02:45 So we can go ahead for now and call that f, but we've also got the first part of our solution right there...

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