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### Course: Digital SAT Math > Unit 2

Lesson 5: Solving systems of linear equations: foundations# Solving systems of linear equations | Lesson

A guide to solving systems of linear equations on the digital SAT

## What are systems of linear equations?

A

**system of linear equations**is usually a set of two linear equations with two variables. and$x+y=5$ are both$2x-y=1$ **linear equations**with two variables.- When considered together, they form a
**system**of linear equations.

A linear equation with two variables has an infinite number of solutions (for example, consider how $(0,5)$ , $(1,4)$ , $(2,3)$ , etc. are all solutions to the equation $x+y=5$ ). However, systems of two linear equations with two variables can have a single solution that satisfies both solutions.

is the$(2,3)$ *only*solution to both$x+y=5$ *and* .$2x-y=1$

In this lesson, we'll:

- Look at two ways to solve systems of linear equations algebraically:
**substitution**and**elimination**. - Look at systems of linear equations graphically to help us understand when systems of linear equations have one solution, no solutions, or infinitely many solutions.
- Explore algebraic methods of identifying the number of solutions that exist for systems with two linear equations.

**You can learn anything. Let's do this!**

## How do I solve systems of linear equations by substitution?

### Systems of equations with substitution

### How does substitution work?

Our goal when solving a system of equations is to reduce

**two equations with two variables**down to**a single equation with one variable**. Since each equation in the system has two variables, one way to reduce the number of variables in an equation is to*substitute*an expression for a variable.Consider the following example:

In a system of equations, both equations are $x$ is equal to $2y$ , the $x$ in the second equation is $2y$ . Therefore, we can plug in $2y$ as a $x$ in the second equation:

*simultaneously true*. In other words, since the first equation tells us that*also*equal to*substitute*forFrom here, we can solve the equation $3y=3$ , then use the value of $y$ to calculate $x$ .

To solve a system of equations using substitution:

- Isolate one of the two variables in one of the equations.
- Substitute the expression that is equal to the isolated variable from Step 1 into the other equation. This should result in a linear equation with only one variable.
- Solve the linear equation for the remaining variable.
- Use the solution of Step 3 to calculate the value of the other variable in the system by using one of the original equations.

#### Let's look at some more examples!

What is the solution $(x,y)$ to the system of equations above?

What is the solution $(x,y)$ to the system of equations above?

### Try it!

## How do I solve systems of linear equations by elimination?

### System of equations with elimination

### How does elimination work?

Our goal when solving a system of equations is to reduce

**two equations with two variables**down to**a single equation with one variable**. Since each equation in the system has two variables, one way to reduce the number of variables is to add or subtract the two equations in the system to cancel out, or*eliminate*, one of the variables.Consider the following system of equations:

Recall that when we're solving equations, we can perform the same operations to both sides of the equation and maintain the equality. Since the second equation tells us that $2x+y$ is equal to $8$ , we can add $2x+y$ to the left side of the first equation, add $8$ to the right side of the first equation, and maintain the equality:

Notice that the $y$ -terms cancel out and are

*eliminated*as a result of adding the two equations. When solving systems of equations using elimination, we're always looking for opportunities to cancel out terms.- If two terms have the
*opposite*coefficients like in the system above ( and$-y$ ), we can$y$ *add*the two equations to cancel the terms. - If two terms have the
*same*coefficients, we can*subtract*the two equations to cancel the terms.

From here, we can solve the equation $5x=15$ , then use the value of $x$ to calculate $y$ .

Sometimes, the system of equations does not have coefficients that readily cancel out. Consider this example:

In this case, we need to find ways to match a pair of coefficients by rewriting one of the equations. There are two ways to do this.

**Option 1:**We can set the system up for eliminating the

*through addition*by multiplying both sides of the first equation by

From here, we can add the second equation to eliminate the $y$ -terms:

**Option 2:**We can also set the system up for eliminating the

*through subtraction*by multiplying both sides of the second equation by

From here, we can subtract the equation from the first equation to eliminate the $x$ -terms:

To solve a system of equations using elimination:

- Identify a pair of terms in the system that have both the same variable and coefficients with the same magnitude (ex:
and$2x$ , or$2x$ and$3y$ ). If necessary, rewrite one or both equations so that a pair of terms have both the same variable and coefficients with the same magnitude.$-3y$ - Add or subtract the two equations in the system to eliminate the terms identified in Step 1. This should result in a linear equation with only one variable.
- Solve the linear equation to obtain a value for the variable.
- Now that you have figured out the value of one variable, plug that value into either equation to find the value of the other variable.

### Let's look at some more examples!

What is the solution $(x,y)$ to the system of equations above?

What is the solution $(x,y)$ to the system of equations above?

### Try it!

## When do I use substitution, and when do I use elimination?

### It's up to you!

All systems of linear equations can be solved with

*either*substitution*or*elimination. On test day, you should use whichever method you're more comfortable with.**Substitution**is sometimes easier when:

- A variable is already isolated:
${x}=4y+1$ - You can isolate a variable in a single step:
$-3x{+y}=7$

**Elimination**is sometimes easier when:

- Both equations contain an identical term:
and${2x}+3y=11$ ${2x}+7y=23$ - The equations contain opposite terms:
and$2x{+2y}=7$ $5x{-2y}=14$ - An equation contains a term that is an integer multiple of a term in the other equation:
and$3x{+4y}=26$ .$5x{+2y}=20$

### Try it!

### Self-reflection

Even though we can solve the systems of equations above using either substitution or elimination, ask yourself these questions:

- For each system, which solution method came to mind first?
- How comfortable am I with isolating variables?
- How comfortable am I with multiplying both sides of an equation by a constant?
- How comfortable am I with adding and subtracting two linear equations?
- How comfortable am I with making more complex substitutions, e.g., substituting for
instead of$5y$ ?$y$ - How comfortable am I with finding least common multiples and using them to set up eliminations?

Your answers to these questions should inform which method you use.

## What is the relationship between lines and the number of solutions to systems of linear equations?

### Linear systems by graphing

### Intersections and number of solutions

A linear equation can be represented by a line in the $xy$ -plane. The

**solution**to a system of linear equations is the point at which the lines representing the linear equations**intersect**.Two lines in the $xy$ -plane can intersect once, never intersect, or completely overlap. Each of these scenarios corresponds to a different

**number of solutions**to the system of equations the two lines represent.- If the two lines have two
*different slopes*, then they will*intersect once*. Therefore, the system of equations has exactly*one solution*. - If the two lines have the
*same slope*but different, then they are parallel lines, and they will -intercepts$y$ *never intersect*. Therefore, we can say that the system of equations has*no solutions*. - If the two lines have the
*same slope*and the*same*, then they will -intercept$y$ *completely overlap*—they are the*same line!*. When this is the case, we say that the system has*infinitely many solutions*.

### Try it!

## How do I determine the number of solutions for systems of linear equations?

### How to determine the number of solutions to a system of equations algebraically

### How do I identify the number of solutions?

In the previous section, we covered the graphical method of determining the number of solutions to a system of linear equations. However, when we don't have the aid of a graph, we can determine the number of solutions algebraically.

One way to do it is to rewrite both equations in slope-intercept form, $y=mx+b$ . This allows us to compare the slopes of the lines, $m$ , and their $y$ -intercepts, $b$ , to determine the number of solutions.

- If the two equations have
*different*, then the system has -values$m$ *one solution*. - If the two equations have
*the same*but -value$m$ *different*, then the system has -values$b$ *no solution*. - If the two equations have both
*the same*and -value$m$ *the same*, then the system has -value$b$ *infinitely many solutions*.

To determine the number of solutions a system of linear equations has using slope-intercept form, $y=mx+b$ :

- Rewrite both equations in slope-intercept form.
- Compare the
- and$m$ -values of the equations to determine the number of solutions.$b$

#### Let's look at an example!

How many solutions does the system of equations above have?

### Try it!

## Your turn!

## Things to remember

To determine the $y=mx+b$ :

**number of solutions**a system of linear equations has using slope-intercept form,- Rewrite both equations in slope-intercept form.
- Compare the
- and$m$ -values of the equations to determine the number of solutions.$b$ - If the two equations have
*different*, then the system has -values$m$ *one solution*. - If the two equations have
*the same*but -value$m$ *different*, then the system has -values$b$ *no solution*. - If the two equations have both
*the same*and -value$m$ *the same*, then the system has -value$b$ *infinitely many solutions*.

## Want to join the conversation?

- Why the English is tough not the math(99 votes)
- I wish I could relate(113 votes)

- why is math so confusing?(44 votes)
- Cause the teacher is confused.(4 votes)

- I don't understand the last question on the page.(14 votes)
- in the last question they told "The system of equations above has no solutions". Then you can know that if a system has no solution the two equations have the same slope. If a and b are the same then a=b and a/b=1(41 votes)

- I truly, for the life of me, cannot understand that final question no matter how hard I try. any tips?(11 votes)
- It was earlier stated that if the slope of both equations are same, and the y-intercept is different, then the lines will be parallel, and if they are parallel they wont intersect, which means no solutions formed.

Now, a and b are the slopes of the first and second equation. The question says no solution, which means their slopes are same but their y-intercepts are different. So if slope of first equation for example is a=1, then slope of second equation b would also be 1. And the question asks a/b, which means 1/1, which is 1.

Now no matter what values you put, like a=2, b will be the same and dividing them would get you 1 as the solution.(27 votes)

- The one and only tip I can give to someone who doesn't understand maths is; keep on staring at the explanation and learn it yourself xdddd not kidding that's how I study, and go back to something you didn't understand because that's the main reason you are getting stuck at some parts and ending up skipping the solution. hope that helps. take your time my friend... wish me luck 333(11 votes)
- True that, maths takes a lot of time to
**truly understand**. Even if one fails just keep on trying and ALWAYS READ THE CORRECTIONS FOR QUESTIONS YOU GET WRONG. U need to know where your mistakes are to avoid making them again! Keep pushing, you have got this!(7 votes)

- is the math portion of the dsat adaptive?(10 votes)
- i think the first module is non-adaptive and the second one is adaptive, but you can still skip around in the second module because the questions are based on how you did in the first one.(6 votes)

- if they have different m but the same b, that would also mean 1 solution right?(5 votes)
- Yes, since they will only intersect once(9 votes)

- I should have done this earlier(9 votes)
- I still didn't understand anything from the last question. Please someone care to explain?(2 votes)
- in the last question they told "The system of equations above has no solutions". Then you can know that if a system has no solution the two equations have the same slope. If a and b are the same then a=b and a/b=1(7 votes)