Solutions to a System of Equations

Section 6.1 Solutions to a System of Equations

Definition 6.1.1.

solution

(x, y)

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Example 6.1.2.

Let’s see if

(3, 5)

is a solution to the equation

y = 2 x - 1 .

In order to check, we will plug in

x = 3

and

y = 5

and see if we get something that’s true:

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\begin{aligned} y & = 2 x - 1 \\ 5 & = 2 (3) - 1 \\ 5 & = 5 \end{aligned}

Indeed,

5 = 5

is a true statement, so

(3, 5)

is a solution to the equation.

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Checkpoint 6.1.3.

Determine if each of the following is a solution to the equation

2 x + y = 2 :

$(2, - 1)$

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$(- 2, 6)$

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$4$

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Answer.

$(2, - 1)$ is not a solution

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$(- 2, 6)$ is a solution

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$4$ is not a solution

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Solution.

Determine if each of the following is a solution to the equation

2 x + y = 2 :

$(2, - 1) :$ In order to determine if this is a solution, we plug in $x = 2$ and $y = - 1 :$

$\begin{aligned} 2 x + y & = 2 \\ 2 (2) + (- 1) & = 2 \\ 3 & = 2 \end{aligned}$

Of course, $3 \neq 2,$ so we ended up with something that is not true. Therefore, $(2, - 1)$ is not a solution to this equation.

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$(- 2, 6) :$ In order to determine if this is a solution, we plug in $x = - 2$ and $y = 6 :$

$\begin{aligned} 2 x + y & = 2 \\ 2 (- 2) + (6) & = 2 \\ 2 & = 2 \end{aligned}$

Of course, $2 = 2,$ so we ended up with something that is true. Therefore, $(- 2, 6)$ is a solution to this equation.

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$4 :$ Remember that a solution is a point, which includes two numbers. This is just a single number, so it cannot be a solution to the equation.
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Definition 6.1.4.

system of equations

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In this class, we will only look at systems of equations that have two equations and two variables. However, in later classes, you might look at more complicated systems that have more than two equations or more than two variables. In that case, you’ll learn more advanced techniques that can handle the more complicated situations.

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Definition 6.1.5.

solution to a system of equations

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Example 6.1.6.

Let’s determine if

(4, 1)

is a solution to the following system of equations:

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{\begin{cases} y & = x - 3 \\ x - 4 y & = 6 \end{cases}

Just as we did before, we will plug in

x = 4

and

y = 1 .

However, this time, we need to plug it into both equations and check that it works for both. Let’s start with the first one:

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\begin{aligned} y & = x - 3 \\ (1) & = (4) - 3 \\ 1 & = 1 \end{aligned}

Now, let’s check the second one:

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\begin{aligned} x - 4 y & = 6 \\ (4) - 4 (1) & = 6 \\ 0 & = 6 \end{aligned}

Since the point only worked for the first equation, but not the second, it is not a solution to this system of equations.

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Checkpoint 6.1.7.

Suppose we have the system of equations:

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

Check if each of the following is a solution to the system:

$(1, - 1)$

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$(2, - 5)$

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$(- 3, 7)$

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Answer.

The point $(1, - 1)$ is not a solution
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The point $(2, - 5)$ is a solution
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The point $(- 3, 7)$ is not a solution
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Solution.

To check if $(1, - 1)$ is a solution, we plug it into both equations. Let’s start with the first equation:
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$\begin{aligned} 3 & = 4 x + y \\ 3 & = 4 (1) + (- 1) \\ 3 & = 4 - 1 \\ 3 & = 3 \end{aligned}$

That worked! Now, let’s check the second equation:
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$\begin{aligned} - 11 & = 2 x + 3 y \\ - 11 & = 2 (1) + 3 (- 1) \\ - 11 & = 2 - 3 \\ - 11 & = - 1 \end{aligned}$

That one did not work. Since the point has to work for both equations in order to be a solution to the system, the point $(1, - 1)$ is not a solution to this system.
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To check if $(2, - 5)$ is a solution, we plug it into both equations. Let’s start with the first equation:
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$\begin{aligned} 3 & = 4 x + y \\ 3 & = 4 (2) + (- 5) \\ 3 & = 8 - 5 \\ 3 & = 3 \end{aligned}$

That worked! Now, let’s check the second equation:
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$\begin{aligned} - 11 & = 2 x + 3 y \\ - 11 & = 2 (2) + 3 (- 5) \\ - 11 & = 4 - 15 \\ - 11 & = - 11 \end{aligned}$

That one also worked! Since it worked for both equations, the point $(2, - 5)$ is a solution to this system.
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To check if $(- 3, 7)$ is a solution, we plug it into both equations. Let’s start with the first equation:
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$\begin{aligned} 3 & = 4 x + y \\ 3 & = 4 (- 3) + (7) \\ 3 & = - 12 + 7 \\ 3 & = - 5 \end{aligned}$

That did not work. At this point, we could plug it into the second equation, but that would end up just being a waste of time. It doesn’t matter if it works for the second one if we already know it doesn’t work for the first one. It has to work for both in order to be a solution, so the point $(- 3, 7)$ is not a solution to this system.
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