# How do we solve a system of linear equations using any method

TLDRThis educational video teaches viewers how to solve a system of linear equations using the method of multiplication by least common multiples. The instructor demonstrates how to find the LCM of the coefficients, multiply the equations accordingly, and then subtract or add the equations to eliminate variables. The process is illustrated with a step-by-step example, showing how to find the values of x and y that satisfy both equations, ultimately leading to the solution of the system.

### Takeaways

- ๐งฎ To solve a system of linear equations, find the least common multiple (LCM) of the coefficients to make them the same.
- ๐ Multiply each equation by a number so that the coefficients of one variable are equal, facilitating its elimination.
- ๐ When choosing which variable to eliminate, consider the LCM of the coefficients of either variable.
- โ๏ธ Apply the distributive property to ensure every term in the equation is multiplied by the chosen multiplier.
- ๐ After creating equivalent equations, subtract one from the other to eliminate one variable.
- ๐ For equations with positive and negative coefficients, adding them will eliminate the variable with the differing signs.
- ๐ข When solving for one variable, use the values from the original or equivalent equations, whichever is simpler.
- ๐ Substitute the value of one variable into an equation to solve for the other variable.
- ๐ The solution to the system of equations is the point of intersection, represented as coordinates (x, y).
- ๐ The process involves algebraic manipulation and strategic choosing of multipliers to simplify the system.

### Q & A

### What is the method described in the transcript for solving a system of linear equations?

-The method described is to find the least common multiple (LCM) of the coefficients and then multiply each equation by the necessary factor to make the coefficients the same, allowing for elimination of one variable by subtracting the equations.

### Why is finding the LCM important when solving a system of linear equations?

-Finding the LCM is important because it helps to make the coefficients of the variable you want to eliminate the same in both equations, which is necessary for the elimination method to work effectively.

### How do you determine the LCM between 7 and 3?

-The LCM between 7 and 3 is 21. To get to 21, you would multiply the top equation by 3 and the bottom equation by 7.

### What is the LCM between 5 and negative 4, and how does it affect the equations?

-The LCM between 5 and 4 (ignoring the negative sign) is 20. This means you would multiply the top equation by 4 and the bottom equation by 5 to align the coefficients of the y variable.

### Why does the sign of the coefficients matter when eliminating variables?

-The sign of the coefficients matters because it determines whether you add or subtract the equations to eliminate a variable. If the coefficients are the same sign, you subtract; if they are opposite signs, you add.

### What is the purpose of multiplying each term in an equation by the LCM multiplier?

-Multiplying each term by the LCM multiplier ensures that you create equivalent equations, which is necessary for the elimination process to be valid.

### What happens when you subtract the two equations after aligning the coefficients?

-When you subtract the two equations after aligning the coefficients, the variable you aimed to eliminate (in this case, x) will cancel out, allowing you to solve for the remaining variable (y in this example).

### How do you find the value of x once you have the value of y?

-Once you have the value of y, you substitute it back into one of the original or equivalent equations to solve for x.

### What is the final solution for the system of equations given in the transcript?

-The final solution is x = -1 and y = -1, which is the coordinate point of intersection for the system of equations.

### Why might one choose to use the second original equation to solve for x instead of the first?

-One might choose to use the second original equation to solve for x because it may have simpler numbers or coefficients that make the calculation easier.

### Outlines

### ๐ Solving Systems of Equations by Elimination

The speaker introduces a method for solving systems of equations using the elimination technique. They explain the importance of finding the least common multiple (LCM) of the coefficients to align the terms that will be eliminated. The example given involves equations with coefficients 7 and 3, and 5 and -4, where the LCMs are 21 and 20, respectively. The speaker chooses to eliminate the x variable by multiplying the first equation by 3 and the second by 7, resulting in new equations. They emphasize the need to apply the distributive property correctly to ensure the equations remain equivalent. The process leads to finding the value of y, which is -1, and then using this value to solve for x, which also turns out to be -1. The solution to the system of equations is the point (-1, -1).

### Mindmap

### Keywords

### ๐กSubstitution

### ๐กElimination

### ๐กLeast Common Multiple (LCM)

### ๐กCoefficients

### ๐กDistributive Property

### ๐กEquivalent Equations

### ๐กVariable

### ๐กMultiplying Equations

### ๐กSolving for a Variable

### ๐กCoordinate Point of Intersection

### Highlights

Introduction to solving systems of linear equations using substitution and elimination methods.

Explanation of the challenge with finding the least common multiple of coefficients in complex equations.

Determining the least common multiple (LCM) between coefficients to facilitate equation manipulation.

Multiplying equations to align coefficients for elimination, specifically using LCM of 7 and 3 which is 21.

Multiplying the top equation by 3 and the bottom equation by 7 to eliminate x's.

Discussing the choice between eliminating x or y and the decision to eliminate x's.

Using distributive property to ensure all terms in equations are multiplied by the chosen multiplier.

Creating equivalent equations by multiplication: 21x + 15y = -36 and 21x - 28y = 7.

Subtracting the equations to eliminate x and solve for y.

Calculation of y after subtracting the equations, resulting in y = -1.

Using the value of y to solve for x by substituting back into one of the original equations.

Solving for x using the second original equation, resulting in x = -1.

Conclusion that the solution to the system of equations is the point (-1, -1).

Emphasizing the importance of choosing the right multipliers to simplify the solving process.

Highlighting the flexibility in choosing which variable to eliminate first.

Advice on using the original equations for solving x and y due to simpler numbers.

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