Systems Of Linear Equations

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Linear equations are equations that describe a straight line. A system of linear equations is a set of two or more linear equations that need to be solved together. In this tutorial, we will learn how to solve systems of linear equations for the SAT exam.

Types of Systems of Linear Equations

There are three types of systems of linear equations: consistent, inconsistent, and dependent.

A consistent system of linear equations is one that has at least one solution. An inconsistent system of linear equations is one that has no solution. A dependent system of linear equations is one that has an infinite number of solutions.

Solving Systems of Linear Equations

To solve a system of linear equations, we use either substitution, elimination or graphing.

Substitution

Substitution is the process of replacing one variable with an expression that involves the other variable. To solve a system of linear equations using substitution, follow these steps:

1. Solve one of the equations for one of the variables in terms of the other variable.
2. Substitute the expression obtained in step 1 into the other equation and solve for the other variable.
3. Substitute the value obtained in step 2 into the expression obtained in step 1 to find the value of the first variable.
4. Check your solution by substituting the values obtained in steps 2 and 3 into both equations.

Let's solve an example system of linear equations using substitution:

$\begin{cases}2x + 3y = 7\\4x - y = 11\end{cases}$

Solve the second equation for y:

$y = 4x - 11$

Substitute this expression into the first equation:

$2x + 3(4x - 11) = 7$

Solve for x:

$2x + 12x - 33 = 7$ $14x = 40$ $x = \frac{20}{7}$

Substitute this value into the expression for y:

$y = 4\left(\frac{20}{7}\right) - 11 = \frac{3}{7}$

Therefore, the solution to the system of linear equations is $(\frac{20}{7}, \frac{3}{7})$.

Elimination

Elimination is the process of adding or subtracting equations to eliminate one of the variables. To solve a system of linear equations using elimination, follow these steps:

1. Multiply one or both of the equations by a constant so that the coefficients of one of the variables are equal and opposite.
2. Add or subtract the equations to eliminate one of the variables.
3. Solve for the remaining variable.
4. Substitute this value into one of the original equations to solve for the other variable.
5. Check your solution by substituting the values obtained in steps 3 and 4 into both equations.

Let's solve the same example system of linear equations using elimination:

$\begin{cases}2x + 3y = 7\\4x - y = 11\end{cases}$

Multiply the second equation by 3:

$\begin{cases}2x + 3y = 7\\12x - 3y = 33\end{cases}$

Add the two equations:

$14x = 40$

Solve for x:

$x = \frac{20}{7}$

Substitute this value into the second equation:

$4\left(\frac{20}{7}\right) - y = 11$

Solve for y:

$y = \frac{3}{7}$

Therefore, the solution to the system of linear equations is $(\frac{20}{7}, \frac{3}{7})$.

Graphing Systems of Linear Equations

Another way to solve systems of linear equations is by graphing the equations on the coordinate plane and finding the intersection point. To graph a linear equation, we first rewrite it in slope-intercept form:

$y = mx + b$

where $m$ is the slope and $b$ is the y-intercept. The slope of a line tells us how steep the line is, and the y-intercept tells us where the line intersects the y-axis.

To graph a system of linear equations, we graph each equation and find the point of intersection. The point of intersection is the solution to the system of linear equations.

Let's graph the example system of linear equations:

$\begin{cases}2x + 3y = 7\\4x - y = 11\end{cases}$

We rewrite each equation in slope-intercept form:

$\begin{cases}y = -\frac{2}{3}x + \frac{7}{3}\\y = 4x - 11\end{cases}$

Then we plot the two lines on the coordinate plane:

The intersection point is $(2.857, 0.429)$ or $(\frac{20}{7}, \frac{3}{7})$, which is the solution to the system of linear equations.

More Examples

Problems

1. Solve the system of linear equations:

$\begin{cases}2x - 3y = 1\\x + 2y = 5\end{cases}$

2. Solve the system of linear equations:

$\begin{cases}3x - 2y = 7\\-x + 4y = 1\end{cases}$

Solutions

Problem 1: Using substitution

Solve the second equation for $x$:

$x = 5 - 2y$

Substitute this expression into the first equation:

$2(5-2y) - 3y = 1$

Solve for $y$:

$10 - 4y - 3y = 10 - 7y = 1$

$-7y = 1 - 10 = -9$

$y = \frac{9}{7}$

Substitute this value into the expression for $x$:

$x = 5 - 2\left(\frac{9}{7}\right) = \frac{5 \times 7}{7} - 2\left(\frac{9}{7}\right) = \frac{17}{7}$

Therefore, the solution to the system of linear equations is $(\frac{17}{7}, \frac{9}{7})$.

Problem 2: Using elimination

Multiply the second equation by 3:

$\begin{cases}3x - 2y = 7\\-3x + 12y = 3\end{cases}$

Add the two equations:

$10y = 10$

Solve for $y$:

$y = 1$

Substitute this value into the second equation:

$-x + 4(1) = 1$

Solve for $x$:

$x = 3$

Therefore, the solution to the system of linear equations is $(3, 1)$.

Conclusion

Solving systems of linear equations is an important topic on the SAT and other standardized tests. There are several methods to solve systems of linear equations, including substitution, elimination, and graphing. It is important to be familiar with all of these methods and know when to use each one.

When solving systems of linear equations, it is also important to check your solution by substituting it into both equations to make sure it satisfies both equations.

By practicing solving systems of linear equations, you can improve your problem-solving skills and increase your chances of success on the SAT and other standardized tests.

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