(5 x)/(2)+3 y=(9)/(2)+y (1)/(4) x-(1)/(2) y=6 x+12 |...

Name: Problem 100, Chapter 4: Intermediate Algebra for College Students
Uploaded: 2021-08-09T17:37:49-08:00
Duration: 5 min 12 s
Channel: Luke P
Description: (5 x)/(2)+3 y=(9)/(2)+y (1)/(4) x-(1)/(2) y=6 x+12

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. The goal is to find the values of the variables that satisfy all the equations at the same time. Techniques such as substitution, elimination, or matrix approaches are commonly used to obtain the solution.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations by combining like terms, isolating variables, and applying arithmetic operations to both sides of the equation. This process is fundamental to solving equations and simplifying expressions.

Clearing Fractions

When equations include fractional coefficients, clearing fractions by multiplying both sides of the equation by the least common denominator can simplify the solving process. This method removes fractions and makes the equation easier to work with.

Substitution and Elimination Methods

These are strategies for solving systems of equations. The substitution method involves expressing one variable in terms of another using one equation and substituting this into the other equation. The elimination method involves adding or subtracting equations to eliminate one variable, thereby reducing the system to a single-variable equation.

Transcript

00:01 So today we are solving system of linear equations with two variables by substitution.

00:08 In this case, this is for problem 19 in this book.

00:12 There is an error.

00:14 So i just want to make sure this is problem 19.

00:19 Okay, so substitution, the way that works is you want to get a variable by itself so that we can substitute it into the other equation.

00:28 Looking here, the first thing i want to do is distribute this to simplify this a bit.

00:35 That gives us 2x plus 2y equals 2 minus y.

00:40 Now it looks like it's pretty easy to get y by itself over here.

00:46 We just have to add 2 to our, excuse me, subtract 2 from both sides and subtract 2.

01:00 From both sides.

01:01 Now what all that will do is that'll cancel out this, and it'll cancel out these, which will give us 2x minus 2 equals minus 3y.

01:15 Now what we want to do is divide by negative 3.

01:21 So we divide by negative 3.

01:23 That gives us y equals 2x minus 2 over negative 3.

01:29 Now we can take that and substitute it in to this equation, which gives us 4x minus 1 equals 2 minus 5 times 2x minus 2 minus 3.

01:51 Now we want to solve 4x.

01:56 We do that by distributing in this negative 5, which gives us 4x minus 1.

02:04 Equals 2 minus 10x plus 10 over negative 3.

02:20 I distributed the negative, so this is going to be plus this.

02:27 Now we want to solve further.

02:31 Let's do that by multiplying both sides by negative 3...

Please give Ace some feedback