# How To Do Substitution In Algebra 1

**How To Do Substitution In Algebra 1** – This is Section 4.2 “Solving Linear Systems by Substitution” in Beginning Algebra (v. 1.0). Click here for details (including license).

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## How To Do Substitution In Algebra 1

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## Solving Linear Systems By Substitution

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### How To Solve Simultaneous Equations Using Substitution Method

This section defines a complete algebraic technique for solving the system. The idea is to solve one equation for one of the variables and substitute the result into another equation. After performing this substitution step, we are left with one equation with one variable that can be solved algebraically. This is called the fallback method. A method of solving linear systems by solving for one of the variables and substituting the result into another equation. , the steps are described in the following example.

Step 1: Solve for one of the variables in the equation. Choosing the first equation, we can isolate

This will give you an equivalent equation in one variable that can be solved using the techniques you have learned so far.

Step 4: Reversely find the value of the substitution variable and substitute it back into one of the original equations or equivalent equation to determine the corresponding value of the other variable. To find the value of another coordinate. reward

## Solving Linear Systems With Substitution

= 1 to the original equation or one of its equivalents. Typically, you use the equivalent equation found when separating the variables in Step 1.

An alternative method for solving systems is the fully algebraic method. So there is no need to describe the line.

The coefficient is 1 in the second equation. This shows that it can be separated in one step as follows:

Answer: (-1, -4). The graph of this particular system is a good exercise in comparing the alternative method to the graph method of solving the system.

#### Solved Solving Systems By Substitution: Part Ii Solve Each

As you know, not all linear systems have only one ordered pair solution. Remember that some systems have infinitely many ordered pair solutions, and some systems have no solutions at all. Next, we investigate what happens when alternative methods are used to solve dependent systems.

Solution: Since the first equation has a term with coefficient 1, we decide to solve it first.

This course led to a true statement. So the equation is identity and any real number is the solution. This indicates that the system is dependent. The format of the concurrent solution is (

To better understand the previous example, write both equations in slope-intercept form and plot them on the same set of axes.

### Algebra 1 Review Systems Of Linear Equations Using Substitution

Since both equations show the same line, we know that the system is dependent. Now investigate what happens when inconsistent systems are resolved in alternative ways.

Solutions lead to incorrect statements. This shows that the equation contradicts itself. no solution

A false statement indicates that the system is inconsistent or geometrically the lines are parallel and do not intersect. To illustrate this, we determine the slope-intercept shape of each line and plot them on the same set of axes.

In the oblique cut form, it is easy to see that the slopes of the two straight lines are the same, but different.

### Solving Systems By Substitution Graphic Organizer

51. The sum of these two numbers is 19. The larger number is 1 less than 3 times the smaller number.

57. At the beginning of a substitution solution, describe what drives you to choose a variable to solve.

Replace it to fix your system. y = 3x y = x – Step 2 Step 1 y = 3x Solve both equations for y. y = x – 2 Step 2 y = x – 2 3x = x – 2 Replace y with 3x in the second equation. Step 3 Solve for –x –x 2x = –2 2x = –2 x = –1 x. Subtract x from both sides then divide by 2.

Write one of the original equations. Step 4 Substitute y = 3x y = 3(-1) y = -3 x -1. Write your solutions in ordered pairs. Choose the substitution (–1, –3) in the two equations of the five-step (–1, –3) system. y = 3x –3 3(–1) –3 –3 y = x – 2 –3 –1 – 2 –3 –3

### Solution: Algebra 1 Equations

Replace it to fix your system. y = x + 1 4x + y = 6 The first equation is solved for y. Step 1 y = x + 1 Step 2 4x + y = 6 4x + (x + 1) = 6 Substitute x + 1 for y in the second equation. 5x + 1 = 6 simplifies. Solve for x. Step 3 –1 –1 5x = 5 x = 1 5x = 5 Subtract 1 from both sides. Divide both sides by 5.

Write one of the original equations. Step 4 y = x + 1 y = 1 + 1 y = 2 Replace x with 1. Write your solutions in ordered pairs. Check the substitution (1, 2) in the two equations of the five-level (1, 2) system. y = x + 1 2 2 4x + y = 6 4 (1) 6 6

Replace it to fix your system. x + 2y = –1 x – y = 5 Step 1 x + 2y = –1 Solve the first equation for x by subtracting 2y from both sides. −2y −2y x = –2y – Step 1 2 x – y = 5 (–2y – 1) – y = 5 Replace x in the second equation with –2y – 1. –3y – 1 = 5 Simplify.

8 Example 1C Step 3 continued –3y – 1 = 5 Solve for y. +1 +1 –3y = 6 Add 1 to both sides. –3y = 6 –3 –3 y = –2 Divide both sides by –3. Step 4 x – y = 5 Write one of the original equations. x – (–2) = 5 x + 2 = 5 Substitute –2 for y. –2 –2 x = 3 Subtract 2 from both sides. Write your solutions in ordered pairs. Level 5 (3, –2)

## Grade 8 Algebra: The Substitution Method] How Do I Do Questions 2 & 3?

9 Find out! Example 1a Solve the system by substitution. y = x + 3 y = 2x + 5 Step 1 y = x + 3 y = 2x + 5 Both equations are solved for y. Step 2 2x + 5 = x + 3 y = x + 3 Replace y with 2x + 5 in the first equation. –x – 5 –x – 5 x = –2 Step 3 Solve for 2x + 5 = x + 3 x. Subtract x and 5 from both sides.

Replace it to fix your system. Write one of the original equations. Step 4 Substitute y = x + 3 y = –2 + 3 y = 1 x -2. Write the 5-step (–2, 1) solution in ordered pairs.

11 Find out! Example 1b Solve the system by substitution. x = 2y – 4 x + 8y = 16 Step 1 Solve the first equation for x = 2y – 4 x. (2y – 4) + 8y = 16 x + 8y = 16 Step 2 Replace x in the second equation with 2y – 4. Step 3 10y – 4 = 16 Simplify. Then solve for y. 10y = 20 Add 4 to both sides. 10y = Divide both sides by 10. y = 2

Replace it to fix your system. Step 4 x + 8y = 16 Write one of the original equations. x + 8(2) = 16 Replace y with 2. x + 16 = 16 simply. x = 0 – 16 –16 Subtract 16 from both sides. Write your solutions in ordered pairs. Step 5 (0, 2)

## Teach Algebra: Substitution & Solving Equations

13 Find out! Example 1c Solve the system by substitution. 2x + y = -4 x + y = -7 Solve the second equation for x by subtracting y from each.

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