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A system of equations is a set of two or more equations. Systems can be solved graphically and algebraically. In this section, one algebraic method, the substitution method, will be applied to solve a system of linear equations.

This is an algebraic method of finding the solutions to a system of equations. It consists of *substituting* an equivalent expression for a variable in one of the equations of the system. Consider an example system of linear equations.
### 1

The first step is to isolate *any* variable in *any* of the equations. For simplicity, in this case the y-variable will be isolated in Equation (I).
### 2

In the equation where the variable was not isolated, substitute the obtained expression for the variable. In this case, 2x+4 will be substituted for y in Equation (II).
Now, Equation (II) only has *one* variable, which is x.
### 3

Solve the equation that contains only one variable. In this case, Equation (II) will be solved for x.
The value of the x-variable, in this case, is 2.
### 4

Since the value of one variable is known, it can be substituted in the equation that has not been considered yet. In this case, x=2 will be substituted in Equation (I).
The value of the y-variable, in this case, is 8. Therefore, the solution to the system of equations, which is the point of intersection of the lines, is (2,8) or x=2, y=8.
If at any step of the method a true statement is obtained, then the lines represented by the equations of the system are coincidental. Therefore the system has infinitely many solutions. Conversely, if at any step a false statement is obtained, then the lines are parallel. In this case, the system has no solution.

${y−4=2x9x+6=3y (I)(II) $

To solve the system by using the Substitution Method, there are four steps to follow.
Isolate One Variable in Any of the Equations

Substitute the Expression

Solve the Equation With One Variable

${y=2x+49x+6=3(2x+4) (I)(II) $

Distr

$(II):$ Distribute 3

${y=2x+49x+6=6x+12 $

SubEqn

$(II):$ LHS−6=RHS−6

${y=2x+49x=6x+6 $

SubEqn

$(II):$ LHS−6x=RHS−6x

${y=2x+43x=6 $

DivEqn

$(II):$ $LHS/3=RHS/3$

${y=2x+4x=2 $

Substitute the Value of the Variable in the Other Equation

${y=-x−1y=2x−13 $

Show Solution

To use the substitution method, we substitute one equation into the other. Here, since both equations are isolated for y, substituting turns into setting the equations equal to each other. Specifically, we'll substitute $y=-x−1$ into the second.
Next, we can solve the above equation for x.
Thus, the x-coordinate of the solution is x=4. Next, we can use this value to find the y-coordinate. To do this, we can substitute x=4 into either equation. We'll arbitrarily choose the first equation.
Thus, y=-5. This means the solution to the system is (4,-5).

The sum of two numbers is 17. One of the numbers is two more than three times the other number. Write a system that represents the given relationships. Then, find the numbers using substitution.

Show Solution

We can use the given information to write two equations. First, we must define our variables. Let the first number be x and the other y. We know that the sum of these numbers is 17. Thus,
To solve this system using substitution, we must substitute one equation into the other. Let's substitute x=3y+2 into x+y=17. This will allow us to then solve for y.
We've found that the y-coordinate of the solution is y=3.75. We can substitute this value into either equation to solve for the corresponding x-value. We'll choose the second equation.
Since x=13.25, the solution to the system is (13.25,3.75).

x+y=17.

We also know that one of the numbers, let's say x, is two more than three times the other number, which is then y. This gives us the equation
x=3y+2.

Together these two equations create the system
x+y=17

Substitute

x=3y+2

3y+2+y=17

SimpTerms

Simplify terms

4y+2=17

SubEqn

LHS−2=RHS−2

4y=15

DivEqn

$LHS/4=RHS/4$

$y=415 $

CalcQuot

Calculate quotient

y=3.75

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