9. Linear and Simultaneous Linear Equations
Many branches of mathematics and science require the solution of algebraic equations. The most fundamental types of equations are:
- Linear equations
- Simultaneous linear equations
The solutions of these equations are discussed in this chapter.
Do you need this chapter?
Here are some questions on the topics in this chapter to consider. If you can do these you may skip this chapter.
Quiz: Equation solving
1. Solve [latex]10-n=3[/latex] for [latex]n[/latex].
a) [latex]n=7[/latex]
b) [latex]n=-7[/latex]
c) [latex]n=13[/latex]
d) [latex]n=-13[/latex]
e) None of the above.
Answer
a)
2. If [latex]3x-11=13[/latex] then [latex]x=[/latex]
a) [latex]3[/latex]
b) [latex]21[/latex]
c) [latex]8[/latex]
d) [latex]6[/latex]
e) None of the above.
Answer
c)
3. Solve [latex]\dfrac{h}{5}-2 =10[/latex]
a) [latex]52[/latex]
b) [latex]60[/latex]
c) [latex]40[/latex]
d) [latex]17[/latex]
e) None of the above.
Answer
b)
4. If [latex]3j-1=19-j[/latex] then [latex]j=[/latex]
a) [latex]6[/latex]
b) [latex]9[/latex]
c) [latex]10[/latex]
d) [latex]5[/latex]
e) None of the above.
Answer
d)
5. If [latex]3\left( a+4 \right)=2\left(2a+1 \right)[/latex] then [latex]a=[/latex]
a) [latex]3[/latex]
b) [latex]10[/latex]
c) [latex]11[/latex]
d) [latex]6[/latex]
e) None of the above.
Answer
b)
6. If [latex]4-\dfrac{2y}{3}=3[/latex] then [latex]y=[/latex]
a) [latex]6[/latex]
b) [latex]6.5[/latex]
c) [latex]1.5[/latex]
d) [latex]18[/latex]
e) None of the above.
Answer
c)
7. Solve the following simultaneous equations:
\begin{align*}
x+y &=8\\
2x-3y &=1
\end{align*}
a) [latex]x=3[/latex], [latex]y=5[/latex]
b) [latex]x=7[/latex], [latex]y=1[/latex]
c) [latex]x=0[/latex], [latex]y=8[/latex]
d) [latex]x=5[/latex], [latex]y=3[/latex]
e) None of the above.
Answer
d)
If you need to review this topic, go to the relevant section in this Chapter.
A brief introductory video on this subject follows. Further details are given in this Chapter.
Solving linear equations (8:29 min)
Solving linear equations (8:29 min) by RMIT University Library Videos (YouTube)
Linear equations
Linear equations have one variable and are the simplest to solve. The following are examples of linear equations:
\begin{align*}
&x-2=4 \\
&3y-4=0 \\
&2z-3=3z+7 \\
&\dfrac{2p}{5}-8=32 \\
&2\left(3x-1\right)=5\left(x+4\right) .\hspace{8cm}
\end{align*}
To solve a linear equation, you try to get the unknown variable on its own. This can be done using inverse operations. Because we are dealing with an equation, anything you do to the left-hand side of the equals sign must also be done to the right-hand side.
Simple linear equations
To solve
[latex]x-2=4[/latex]
we add [latex]2[/latex] to both sides to get
[latex]x-2+2=4+2.[/latex]
Then simplifying each side,
\begin{align*}
x&=4+2 \\
&=6.
\end{align*}
Note the use of the inverse operation. We wanted to remove negative [latex]2[/latex] from the left-hand side, so we added [latex]2[/latex] to both sides.
Suppose we want to solve [latex]3y=9[/latex]. In this case, the variable [latex]y[/latex] is multiplied by [latex]3[/latex]. To get [latex]y[/latex] by itself we divide both sides by [latex]3[/latex] to get
\begin{align*}
\frac{3y}{3}&=\frac{9}{3}
\end{align*}
and cancelling the threes on the left-hand side we get the solution [latex]y=3[/latex].
More complex cases involve more steps. The order of the steps is not important but it is usually easiest if you add or subtract terms before multiplication or division. For example, if we want to solve
\begin{align*}
3x-2=4 \\
\end{align*}
we first get the [latex]3x[/latex] term on its own. To do this we add [latex]2[/latex] to both sides:
\begin{align*}
3x-2+2&=4+2 \\
3x&=6.
\end{align*}
Now we get [latex]x[/latex] on its own by dividing each side by [latex]3[/latex] to get
\begin{align*}
\frac{3x}{3}&=\frac{6}{3}\\
x&=2.
\end{align*}
In this example, we added [latex]2[/latex] first. If we first divided by [latex]3[/latex], we would have:
\begin{align*}
\frac{3x}{3}-\frac{2}{3}=\frac{4}{3} \\
x&=\frac{4}{3}+\frac{2}{3}\\
&=\frac{6}{3}\\
&=2.
\end{align*}
While dividing by [latex]3[/latex] first gives the same answer, it introduces fractions. Hence it is usually easiest to add terms to get the term containing the variable on its own as the first step.
A linear equation may have the variable on both sides of the equal sign. In this case, we must gather all terms involving the variable to one side of the equal sign. For example to solve
\begin{align*}
p-9&=3-2p
\end{align*}
we add [latex]2p[/latex] to both sides:
\begin{align*}
p-9+2p&=3-2p+2p\\
3p-9&=3.
\end{align*}
Now add [latex]9[/latex] to both sides to get [latex]3p[/latex] on its own.
\begin{align*}
3p-9+9&=3+9\\
3p&=12.
\end{align*}
Finally, divide both sides by [latex]3[/latex] to get the answer:
\begin{align*}
\frac{3p}{3}&=\frac{12}{3}\\
p&=4.
\end{align*}
Examples: Simple linear equations
1. Solve the equation [latex]x-3=6[/latex].
Solution:
We get [latex]x[/latex] on its own by adding [latex]3[/latex]. Adding [latex]3[/latex] to both sides gives:
\begin{align*}
x-3+3&=6+3\\
\implies x&=9.
\end{align*}
Hence the solution is [latex]x=9[/latex].
2. Solve the equation [latex]y+7=12[/latex].
Solution:
We get [latex]y[/latex] on its own by subtracting [latex]7[/latex]. Subtracting [latex]7[/latex] from both sides gives:
\begin{align*}
y+7-7&=12-7\\
\implies y&=5.
\end{align*}
Hence the solution is [latex]y=5[/latex].
3. Solve the equation [latex]3-y=4.[/latex]
Solution:
Subtract [latex]3[/latex] from both sides
\begin{align*}
3-y-3&=4-3\\
\implies -y&=1.
\end{align*}
Now multiply both sides by [latex]-1[/latex] to get a positive [latex]y[/latex] on the left-hand side.
\begin{align*}
-y\times \left(-1\right)&=1\times \left(-1\right)\\
\implies y&=-1.
\end{align*}
Hence the solution is [latex]y=-1[/latex]. Note that [latex]-y=1[/latex] is a valid solution but it looks better to write [latex]y=-1[/latex].
4. Solve the equation [latex]3m=18[/latex].
Solution:
We get [latex]m[/latex] on its own by dividing by [latex]3[/latex]. Dividing both sides by [latex]3[/latex] gives:
\begin{align*}
\frac{3m}{3}&=\frac{18}{3}\\
\implies m&=6.
\end{align*}
Hence the solution is [latex]m=6[/latex].
5. Solve the equation [latex]\dfrac{x}{5}=4.[/latex]
Solution:
We get [latex]x[/latex] on its own by multiplying by [latex]5[/latex]. Multiplying both sides by [latex]5[/latex] gives:
\begin{align*}
\frac{x}{5}\times 5&=4 \times 5\\
\implies x&=20.
\end{align*}
Hence the solution is [latex]x=20[/latex].
6. Solve the equation [latex]\dfrac{3x}{4}=10[/latex].
Solution:
We get [latex]x[/latex] on its own by multiplying by [latex]4[/latex] and then dividing by [latex]3[/latex].
Multiplying both sides by [latex]4[/latex] gives:
\begin{align*}
\frac{3x}{4}\times 4&=10 \times 4\\
\implies 3x&=40.
\end{align*}
Now divide both sides by [latex]3[/latex]:
\begin{align*}
\frac{3x}{3}&=\frac{40}{3}\\
\implies x&=\frac{40}{3}.
\end{align*}
Hence the solution is [latex]x=\frac{40}{3}[/latex].
7. Solve the equation [latex]\dfrac{2p}{3}-4=10[/latex].
Solution:
We first get rid of the [latex]-4[/latex] on the left-hand side. Then we multiply by [latex]3[/latex] and divide by [latex]4[/latex].
Adding [latex]4[/latex] to both sides:
\begin{align*}
\frac{2p}{3}-4+4&=10+4\\
\implies \frac{2p}{3}&=14.\\
\end{align*}
Multiplying both sides by [latex]3[/latex]:
\begin{align*}
\frac{2p}{3}\times 3&=14\times 3\\
\implies 2p&=42\\
\end{align*}
Dividing both sides by [latex]2[/latex]:
\begin{align*}
\frac{2p}{2}&=\frac{42}{2}\\
\implies p&=21.\\
\end{align*}
8. Solve [latex]2p+4=3p-5[/latex]
Solution:
Subtract 2p to get the term contain p to be on one side (the right-hand side):
\begin{align*}
2p+4-2p&=3p-5-2p\\
\implies 4&=p-5\\
\end{align*}
Add [latex]5[/latex] to both sides
\begin{align*}
4+5&=p-5+5\\
\implies 9&=p.\\
\end{align*}
Hence the solution is [latex]p=9[/latex].
Here are some exercises to try.
Exercises: Simple linear equations
Solve the following equations:
1. [latex]x+4=21[/latex]
2. [latex]4-y=3[/latex]
3. [latex]3m=8[/latex]
4. [latex]\dfrac{3p}{2}=6[/latex]
5. [latex]4-\dfrac{5p}{2}=6[/latex]
6. [latex]x+1=3x-6[/latex].
Answers
[latex]\begin{array}{lllllllllllllll} 1.\;x=17 & & 2.\;y=1 & & 3.\;m=8/3 & & \\ 4.\;p=4 & & 5.\;p=-\dfrac{4}{5} & & 6.\;x=\dfrac{7}{2}\end{array}[/latex]
Equations involving square roots or brackets
Equations may involve square roots. For example
\begin{align*}
\sqrt{x-2}&=4.
\end{align*}
In this case, we first remove the square root sign by squaring both sides of the equation:
\begin{align*}
{\sqrt{x-2}}^2&=4^2\\
x-2&=16.
\end{align*}
Now we add [latex]2[/latex] to both sides to get the answer:
\begin{align*}
x-2+2&=16+2\\
x&=18.
\end{align*}
When solving equations with brackets, it is usually a good idea to remove the brackets first. For example, to solve
\begin{align*}
3\left(m-4\right)&=2\left(3m+9\right)\\
\end{align*}
first expand out the brackets:
\begin{align*}
3m-12&=6m+18.
\end{align*}
Next get the terms involving [latex]m[/latex] on one side of the equal sign. Here we subtract [latex]3m[/latex] from both sides:
\begin{align*}
3m-12-3m&=6m+18-3m\\
-12&=3m+18.
\end{align*}
Rearrange to get [latex]3m[/latex] on the left side:
\begin{align*}
3m+18&=-12.
\end{align*}
Subtract [latex]18[/latex] from both sides to get [latex]3m[/latex] on its own:
\begin{align*}
3m+18-18&=-12-18\\
3m&=-30.
\end{align*}
Finally, divide both sides by [latex]3[/latex] to get the solution:
\begin{align*}
3m&=-30\\
\frac{3m}{3}=-\frac{30}{3}\\
m&=-10.
\end{align*}
Examples: Equations with brackets and square roots
1. Solve the equation [latex]4\left(2m-3\right)=\left(m+1\right)[/latex].
Solution:
Removing the brackets we have
\begin{align*}
8m-12&=m+1.
\end{align*}
Subtracting [latex]m[/latex] from both sides,
\begin{align*}
8m-12-m&=m-m+1\\
7m-12&=1.\\
\end{align*}
Adding [latex]12[/latex] to both sides gives
\begin{align*}
7m-12+12&=1+12\\
7m&=13.\\
\end{align*}
Dividing by [latex]7[/latex] gives the solution [latex]x=\dfrac{13}{7}[/latex].
2. Solve the equation [latex]\sqrt{x-1}=4[/latex].
Solution:
Squaring both sides:
\begin{align*}
x-1&=4^2\\
\implies x-1&=16.
\end{align*}
Add [latex]1[/latex] to both sides to get the solution [latex]x=17[/latex].
3. Solve [latex]\sqrt{x^2-1}=x-1[/latex].
Solution:
Squaring both sides gives
\begin{align*}
x^2-1&=\left(x-1\right)^2\\
&=x^2-2x+1.
\end{align*}
Subtracting [latex]x^2[/latex] from both sides,
\begin{align*}
x^2-1-x^2&=x^2-2x+1-x^2\\
\implies -1&=-2x+1.
\end{align*}
Adding [latex]2x[/latex] to both sides gives:
\begin{align*}
-1+2x&=-2x+1+2x\\
\implies -1+2x&=1.
\end{align*}
Adding [latex]1[/latex] to both sides we obtain,
\begin{align*}
-1+2x+1&=1+1\\
\implies 2x&=2.
\end{align*}
Dividing both sides by [latex]2[/latex] gives the solution [latex]x=1[/latex].
Here are some exercises to try.
Exercises: Equations with brackets and square roots
Solve the following equations:
1. [latex]2\left(x-1\right)=3x[/latex]
2. [latex]3 =\sqrt{x-1}[/latex]
3. [latex]2\sqrt{3m-9}=3\sqrt{m+1}[/latex]
Answers
[latex]\begin{array}{lllllllllllllll} \mathbf{1.}\ x=-2 & & \mathbf{2.}\ x=10 & & \mathbf{3.}\ m=15 \end{array}[/latex]
Simultaneous linear equations
Sometimes it is necessary to solve two linear equations simultaneously. There are two ways to do this: graphically or algebraically.
Graphical solution of two linear equations
Suppose we want to solve the two equations
\begin{align*}
2x+3y&=8\qquad (9.1)\\
-x+y&=1. \qquad (9.2)\\
\end{align*}
This is called solving simultaneous equations because the solution is common to both equations. The graph of each of these equations is a straight line. If we graph these equations, the point where they intersect is the solution of the simultaneous equations.
To graph eqn.(9.1) we set [latex]y=0[/latex] to get the [latex]x-[/latex] intercept and [latex]x=0[/latex] to find the [latex]y-[/latex]intercept. Let [latex]y=0[/latex] then eqn.(9.1) becomes:
\begin{align*}
2x&=8\\
x&=4 \quad \text{ after dividing both sides by 2}
\end{align*}
hence the graph passes through [latex]\left(4,0\right)[/latex]. If [latex]x=0[/latex], eqn.(9.1) becomes
\begin{align*}
3y&=8\\
y&=\frac{8}{3} \quad \text{ after dividing both sides by 3}
\end{align*}
hence the graph passes through [latex]\left(0,\frac{8}{3}\right)[/latex]. Plotting these two points and ruling a straight line through them gives the graph in Fig.(9.1)
Now we graph eqn.(9.2). Set [latex]y=0[/latex] then eqn.(9.2) becomes:
\begin{align*}
-x&=1\\
x&=-1 \quad \text{ after multiplying both sides by -1}
\end{align*}
hence the graph passes through [latex]\left(-1,0\right)[/latex]. If [latex]x=0[/latex], eqn.(9.2) becomes
\begin{align*}
y&=1\\
\end{align*}
hence the graph passes through [latex]\left(0,1\right)[/latex]. Figure 9.2 shows the graphs of equations (9.1) and (9.2) on the same axes.
The graphs intersect at the point [latex]\left(1,2\right)[/latex] hence [latex]x=1[/latex] and [latex]y=2[/latex] is the solution to the simultaneous equations.
It is easy to check if the solution is correct. We substitute the calculated solution into the original equations. From eqn.(9.1),
\begin{align*}
2x+3y&=2\times1 + 3\times 2\\
&=2+6\\
&=8
\end{align*}
which is true and, from eqn.(9.2),
\begin{align*}
-x+y&=-1+2\\
&=1
\end{align*}
which is also true. Hence the solution of the equations is [latex]x=1[/latex] and [latex]y=2[/latex].
It is good practice to always check your answer in this way.
The graphical method can suffer from inaccuracy because the solution can only be as accurate as the graph. For increased accuracy, we use an algebraic approach.
Algebraic solution of two linear equations
Algebraic solutions are obtained by eliminating one of the variables to get an equation in one variable. This is solved and the solution is substituted into either of the original equations to get the other variable.
The Elimination method
Consider the simultaneous equations below. They are the same as those solved graphically in the previous section.
\begin{align*}
2x+3y&=8\qquad (9.1)\\
-x+y&=1. \qquad (9.2)\\
\end{align*}
Remember that you can multiply both sides of an equation by any number without changing its meaning. Multiply eqn.(9.2) by [latex]2[/latex] to get
\begin{align*}
-2x+2y&=2. \qquad (9.3)\\
\end{align*}
Note that the left-hand sides of equations (9.1) and (9.3) contain [latex]2x[/latex] and [latex]-2x[/latex] respectively. Now add equations (9.1) and (9.3):
\begin{align*}
2x+3y-2x+2y&=8+2\\
\implies 5y&=10\\
\end{align*}
Note we have eliminated the [latex]x[/latex] variable and can now solve for [latex]y[/latex]:
\begin{align*}
5y&=10 \\
\implies y&=2.
\end{align*}
Now substitute this result into eqn.(9.2)
\begin{align*}
-x+2&=1\\
\implies x&=1.
\end{align*}
Hence we obtain [latex]x=1[/latex] and [latex]y=2[/latex] as the solution, just as we did using the graphical method.
The Substitution method
Consider again the set of equations
\begin{align*}
2x+3y&=8\qquad (9.1)\\
-x+y&=1. \qquad (9.2)\\
\end{align*}
Solving eqn.(9.2) for [latex]y[/latex] by adding [latex]x[/latex] to both sides:
\begin{align*}
-x+y+x&=x+1\\
\implies y&=x+1. \qquad (9.4)\\
\end{align*}
Now substitute eqn.(9.4) into eqn.(9.1):
\begin{align*}
2x+3\left(x+1\right)&=8\\
2x+3x+3&=8.\\
\end{align*}
We now have a single equation in one unknown that can be solved:
\begin{align*}
2x+3x+3&=8\\
\implies 5x+3-3&=8-3\\
\implies 5x&=5\\
\implies x&=1.
\end{align*}
Now substitute [latex]x=1[/latex] into eqn.(9.4):
\begin{align*}
y&=x+1\\
&=1+1\\
&=2.
\end{align*}
Hence the solution is [latex]x=1[/latex] and [latex]y=2[/latex] as before.
Unless you are told which method, (graphical, elimination or substitution) to use, you may use any method to get a solution. However, sometimes the equations themselves suggest the best method as shown in the following examples.
Examples: Simultaneous linear equations
1. Solve the equations
\begin{align*}
2x+3y&=8 \qquad (9.5) \\
y&=2x. \qquad (9.6) \\
\end{align*}
Solution:
In this case, eqn.(9.6) gives [latex]y[/latex] in terms of [latex]x[/latex]. Hence we can substitute for [latex]y[/latex] in eqn.(9.5) to get
\begin{align*}
2x+3\times 2x&=8 \\
\implies 2x+6x&=8\\
\implies 8x&=8\\
\implies x&=1.
\end{align*}
Now substitute [latex]x=1[/latex] into eqn.(9.6) to get [latex]y[/latex]:
\begin{align*}
y&=2\times 1\\
&=2.
\end{align*}
Hence the solution is [latex]x=1[/latex] and [latex]y=2[/latex].
2. Solve the equations
\begin{align*}
2x+3y&=13\qquad (9.7)\\
5x-3y&=1. \qquad (9.8)\\
\end{align*}
Solution:
Note that the first equation contains [latex]3y[/latex] and the second contains [latex]-3y[/latex]. Hence we may remove the [latex]y[/latex] terms by adding the equations:
\begin{align*}
2x+3y+5x-3y&=13+1 \\
\implies 7x&=14 \\
\implies x&=\frac{14}{7}\\
&=2.
\end{align*}
Now substitute [latex]x=2[/latex] into either equation (9.7) or (9.8). It doesn't matter which one you use. We will use eqn.(9.7). Substituting we get
\begin{align*}
2\times 2+3y&=13\\
\implies 4+3y&=13 \\
\implies 4+3y-4&=13-4 \\
\implies y&=\frac{9}{3} \\
&=3.
\end{align*}
Hence the solution is [latex]x=2[/latex] and [latex]y=3[/latex]
3. Solve the equations
\begin{align*}
2x-3y&=9\qquad (9.9)\\
2x+5y&=1. \qquad (9.10)\\
\end{align*}
Solution:
The term [latex]2x[/latex] appears in both equations. Hence subtracting the equations will remove the [latex]x[/latex] terms. You can subtract eqn.(9.9) from eqn.(9.10) or subtract eqn.(9.10) from eqn.(9.9). It doesn't matter as either will remove the [latex]x[/latex] terms. We subtract eqn.(9.10) from eqn.(9.9) to get:
\begin{align*}
2x-3y-\left(2x+5y\right)&=9-1\\
\implies 2x-3y-2x-5y&=8\\
\implies -3y-5y&=8\\
\implies -8y&=8\\
\implies y&=\frac{8}{-8}\\
&=-1.\end{align*}
Now substitute [latex]y=-1[/latex] into eqn.(9.9) or (9.10). We use eqn.(9.9):
\begin{align*}
2x-3\times\left(-1\right)&=9\\
\implies 2x+3&=9\\
\implies 2x&=9-3\\
\implies 2x&=6\\
\implies x&=\frac{6}{2}\\
&=3.
\end{align*}
Hence the solution is [latex]x=3[/latex] and [latex]y=-1[/latex].
4. Solve the equations
\begin{align*}
2x+3y&=4 \qquad (9.11)\\
3x-2y&=-\frac{1}{2}. \qquad (9.12)\\
\end{align*}
Solution:
In this case, there is no clear way forward. Adding or subtracting the equation will not remove one of the variables. However, we are free to multiply each equation by any number. Suppose we want to remove the [latex]x[/latex] terms first. Then we have to get the same coefficient of [latex]x[/latex] in each equation. This can be done by multiplying eqn.(9.11) by [latex]3[/latex] and eqn.(9.12) by [latex]2[/latex] to get
\begin{align*}
6x+9y&=12 \qquad (9.13)\\
6x-4y&=-1. \qquad (9.14)\\
\end{align*}
Subtracting eqn.(9.14) from eqn.(9.13) removes the [latex]x[/latex] terms:
\begin{align*}
6x+9y-\left(6x-4y\right)&=12-\left(-1\right)\\
\implies 13y&=13\\
\implies y&=\frac{13}{13}\\
&=1.
\end{align*}
We obtain [latex]x[/latex] by substituting [latex]y=1[/latex] into either equation. We use eqn.(9.11):
\begin{align*}2x+3\times 1&=4\\
\implies 2x&=4-3\\
&=1\\
\implies x&=\frac{1}{2}.
\end{align*}
Hence the solution is [latex]x=\dfrac{1}{2}[/latex] and [latex]y=1[/latex].
Alternate Solution:
It does not matter which variable is solved for first. Suppose in example 4 above we wanted to solve for [latex]x[/latex] first. The [latex]y[/latex] coefficients are made equal by multiplying eqn.(9.11) by [latex]2[/latex] and eqn.(9.12) by [latex]3[/latex] to get
\begin{align*}
4x+6y&=8\\
9x-6y&=-\frac{3}{2}. \\
\end{align*}
Adding the two equations removes the [latex]y[/latex] terms and allows us to solve for [latex]x[/latex]:
\begin{align*}
4x+6y+9x-6y&=8-\frac{3}{2}\\
\implies 13x&=8-\frac{3}{2}\\
&=\frac{13}{2}\\
\implies x&=\frac{1}{2}.\\
\end{align*}
We find [latex]y[/latex] by substituting [latex]x=\dfrac{1}{2}[/latex] into either equation. Using eqn.(9.11),
\begin{align*}
2\times \frac{1}{2}+3y&=4 \\
\implies3y&=3 \\
\implies y&=1
\end{align*}
as before.
Here are some exercises to try.
Exercises: Simultaneous linear equations
1. Solve the equations:
\begin{align*}
x&=2y\\
3x+3y&=18.
\end{align*}
2. Solve the equations
\begin{align*}
2x+y&=4\\
3x- y&=11. \\
\end{align*}
3. Solve the equations
\begin{align*}
5x+3y&=11 \\
3x-2y&=-1\\
\end{align*}
Answers
[latex]\begin{array}{lllllllllllllll} \mathbf{1.}\ x=4,\ y=2 & & \mathbf{2.}\ x=3,\ y=-2 & & \mathbf{3.}\ x=1,\ y=2\end{array}[/latex]
Key takeaways
- A linear equation has one variable.
- To solve a linear equation use a series of operations on each side of the equation to get the variable on its own.
- Simultaneous linear equations have two variables (they may have more but we consider only two).
- Simultaneous solutions may be solved graphically or algebraically.
- For the graphical method, the solution is the intersection of the graphs of each of the linear equations.
- An algebraic solution involves removing one variable using elimination or substitution. The resulting equation is then solved for the first variable. This is substituted back into either of the two simultaneous equations to solve for the second variable.
