Solve a system of equations by elimination (with decimals) | bprp algebra 2, intermediate algebra

TL;DR

This video explains how to solve a system of equations using the elimination method and provides an example using given equations.

Transcript

we are going to solve this system of equations we have x plus y is equal to 45 and we have 0.2 x plus 0.5 y it's equal to 0.4 times 45. let's use the elimination method because in my opinion about 90 of the time it will be quicker and let's get rid of the x okay so this is one and this is 0.2 on top of this right here we have to get the lowest comm... Read More

Key Insights

• ❓ The elimination method is a useful technique to solve systems of equations efficiently.
• 😘 The lowest common multiple of coefficients is used to manipulate equations and eliminate a variable.
• 🎁 The solution should be presented clearly, providing context for the variables and their meanings.

Questions & Answers

Q: What method does the video recommend for solving the system of equations?

The video recommends using the elimination method because it is typically quicker and more efficient.

Q: How do you eliminate a variable in the system of equations?

To eliminate a variable, you find the lowest common multiple of the coefficients and manipulate one equation to have the same coefficient (with the opposite sign) as the other equation.

Q: What does the value of 'x' represent in the solution?

In the given context, 'x' represents the amount of milliliters of a 20% solution.

Q: How is the final solution presented?

The solution is presented as 'x = 15' and 'y = 30', where 'x' represents the amount of a 20% solution in milliliters, and 'y' represents the amount of a 50% solution in milliliters.

Summary & Key Takeaways

• The video demonstrates how to solve a system of equations by eliminating one variable.

• It explains the process of finding the lowest common multiple and manipulating the equations to eliminate one variable.

• The solution is then obtained by solving for the remaining variable and substituting it back into one of the original equations.