Solve each system of equations. If the system has no...

Solve each system of equations. If the system has no solution, say that it is inconsistent.
$\left\{\begin{array}{l}\frac{1}{3} x-\frac{3}{2} y=-5 \\ \frac{3}{4} x+\frac{1}{3} y=11\end{array}\right.$

Key Concepts

Inconsistent System

An inconsistent system refers to a set of equations that have no common solution. This typically occurs when the equations represent parallel lines that never intersect or if the manipulation of the equations leads to a contradiction. Recognizing an inconsistent system is key to concluding that no solution exists for the given problem.

Clearing Fractions

Clearing fractions is an important algebraic strategy used to simplify equations that contain fractional coefficients. By multiplying both sides of the equation by the least common denominator, the equation is transformed into an equivalent form without fractions, which simplifies the algebraic manipulation and reduces the risk of calculation errors.

System of Linear Equations

A system of linear equations consists of two or more equations that involve the same set of variables. The goal is to find values for the variables that make all the equations true simultaneously. This is fundamental in algebra and is frequently used to model and solve real-world problems where multiple constraints must be satisfied.

Elimination Method

The elimination method is a systematic technique for solving systems of linear equations. It involves adding or subtracting the equations, often after multiplying them by appropriate factors, to cancel out one variable. Once a variable is eliminated, the resulting single-variable equation can be solved, and its value substituted back to obtain the other variable.

Transcript

00:01 To solve this system of equations, i will be using the elimination method.

00:06 Now, this problem looks a little bit tricky because of all the fractions, and they all have different denominators.

00:12 Okay, my advice to you would be to get rid of some of those fractions, and we could do that while we're looking for opposite coefficients.

00:21 So what i'm going to choose to do is multiply the top equation by two.

00:27 And i'm doing that because i'm going to try and get opposite coefficients with my y variable because one's already negative and one's already positive.

00:37 Okay, remember when you multiply by two, you have to distribute that to each term in your equation.

00:47 So let's rewrite this off to the side.

00:49 One third x times two is two thirds x.

00:56 Three over two y times two is three y.

01:02 And negative five times two is negative 10.

01:07 Okay, so since this came out to be negative three, three, three.

01:10 I want my other equation to have a positive 3y.

01:16 So looking at that, i'm also going to have to multiply this equation.

01:20 Now, how can i get one -third to say a positive three? well, think through it.

01:27 If you were to multiply it by nine, that might do the trick.

01:33 Again, multiplied each term in your equation.

01:38 So three -fourths times nine, you're really just doing three times nine.