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Solving linear equations

Earlier, we looked at how to transpose formulas. Now, let's combine those skills with arithmetic to solve linear equations. By mastering the techniques for solving these equations, you'll be equipped to tackle a wide range of problems and build a strong basis for more advanced mathematical concepts.

Video tutorial – solving linear equations

Watch this video to learn how to solve linear equations.

Solving equations using subtraction; let’s look at a simple equation, X plus three equals seven. So we’re saying what number instead of X which, when added to three, gives us seven. Well obviously the answer is four because four plus three equals seven. An equation is like a balance, the numbers on the left hand side must balance the numbers on the right hand side. But let’s do this another way, let’s get X by itself, in other words we’re doing it mathematically. To do this we subtract the three from the left hand side to isolate the X, so now the left hand side is X plus three minus three which gives us X. Let’s bring in the right hand side, but remember we have to keep the equation balanced, in other words X plus three minus three on the left hand side is equal to seven minus three on the right hand side. You see how the minus three undoes the plus three, so on the left hand side we’ve got X and the right hand side we’ve got four. Notice that the three has been subtracted from both sides, in other words what we do to one side we must do to the other.

So the rule is that the opposite or inverse operation to plus is minus. In other words subtracting undoes adding. Let’s solve an equation using addition. So we’ll start with Y minus six equals 13, so we’re saying what number instead of Y which when added to minus six gives us 13. Well the answer is 19 because 19 take away six equals 13. Again notice the balancing act, 19 minus six on the left hand side is equal to 13 on the right hand side. But let’s do this mathematically. Let’s get Y by itself and to do this we need to add the six from the left hand side to isolate the Y, remember we had to subtract the number with the previous problem, so the left hand side is now Y minus six plus six, which gives us Y, that’s the left hand side. What about the right hand side. Well what we do to one side we do to the other so Y minus six plus six is equal to 13 plus six, notice plus six undoes minus six. In other words Y is equal to 19. Notice again that the six has been added to both sides, so the rule here is that the opposite or inverse operation to minus is plus, in other words adding undoes subtracting.

Let’s solve an equation using division. Let’s look at the equation four K equals 12. Remember four K means four times K, so we’re saying what number instead of K which when multiplied by four gives us 12, well three because four times three is equal to 12, so we have 12 equals 12, our balancing act again. But let’s do this another more mathematical way, let’s get that K by itself and to do this we divide by four on the left hand side to isolate the K. To balance this we must do the same thing on the right hand side, so here we have the left hand side which is four K divided by four, the right hand side we must divide that by four too, so it’s 12 divided by four, there’s the calculation where we’re dividing by four and so we’ve got 12 divided by four is equal to three. Notice this time I’ve just checked the solution to this equation, in other words if K equals three and we substitute four K into the original equation, four times three is equal to 12. So the rule here is that the opposite or inverse operation to multiplication is division. In other words dividing undoes multiplying.

Let’s solve an equation using multiplication. Now here we have an equation M divided by four is equal to minus five, in other words what number instead of M which when divided by four gives us minus five. This one isn’t so easy to work out so we have to use the following rule, in other words let’s use a little bit of mathematics. What we’re doing here is multiplying both sides by four, so M over four times four over one, notice how the fours cancel, is equal to minus five times four over one, or four. In other words minus five times four is minus 20. So let’s check the solution. If the answer is minus 20 then minus 20 divided by four is equal to minus five. So the rule here, for multiplication, is that the opposite or inverse operation to division is multiplication, multiplying undoes dividing.

Now let’s move on to more complicated operations where we have more than one. For instance, let’s look at the equation two P plus five equals seven. The unknown value P is multiplied by two and then five is added to obtain the answer of seven. To find P we undo two P plus five in the reverse order to which it was built up. In other words we subtract five first and then divide by two, which is the opposite to what we did in getting from two P plus five to seven. So let’s subtract five first, two P plus five minus five equals seven minus five, which gives us two. Then let’s divide by two, notice we’re doing it on both sides of the equation, two times P over two equals two over two, the right hand side is now one because we cancelled the two top and bottom, so two P equals two then P equals one.

So let’s look at another equation, but this time it’s a division equation for starters, three X minus one divided by two is equal to one. In order to solve the equation the operations is to build up the equation must be undone in the reserve order. In other words we start by dividing by two and then we add the one. Let’s look at this in more detail. Three X minus one divided by two, let’s multiply both sides by two to get rid of the two on the bottom of the fraction on the left hand side so we end up with three X minus one equals two. Now let’s get rid of the minus one on the left hand side, so three X minus one plus one equals two plus one, notice we’re doing the same operation on both left and right hand sides of the equation. That leaves us with three X equals three here, so the final step is to remove the three on the left hand side and because we have three times X we must divide by three so three X over three equals three over three. Notice the three cancels out and we’re left with X equals one.

So when we solve equations using more than one operation we first of all look at how the operation was built up, then we solve the equation by undoing it in the reverse order to which it was built up. Now go to the exercise sheet and have a go at the problems for yourself. Thank you.

Three examples of linear equations. The first equation is 3 times x plus 1 equals 7. The second equation is 2y over 5 minus 3 equals 1. The third equation is h plus 1 over by 2 equals h minus 2 over by 3.

Linear equations

A linear equation has one unknown variable and may be solved by transposing the equation to make the variable the subject of the equation. For example, the equation:

\[\begin{align*} 3(2x-7) & =-15
\end{align*}\]

may be transposed to get:

\[\begin{align*} x & =6
\end{align*}\]

Transposition involves changing the equation via operations like addition, subtraction, multiplication and division. Let's look at how to solve simple and more complex linear equations.

Solving simple equations

Remember that when you transpose formulas, whatever you do to one side, you must do to the other.

If you add, subtract, multiply or divide one side of the equation by a number, you must do the same to the other side. This keeps the equation balanced, or preserves equality.

Example 1 – solving simple equations

Solve \(m+4=-2\).

We need to make the variable, \(m\), the subject of the equation. \(4\) is being added to \(m\), so we need to subtract \(4\) from both sides.

\[\begin{align*} m+4 & = -2\quad\textrm{subtract \(4\) from both sides}\\
m+4-4 & = -2-4\\
m & = -6
\end{align*}\]

The solution is \(m=-6.\) You can always check your solution by substituting your answer into the original equation and making sure the left-hand side (LHS) of the equation is equal to the right-hand side (RHS).

\[\begin{align*} \textrm{LHS} & = (-6)+4\\
& = -2\\
& = \textrm{RHS}
\end{align*}\]

Solve \(p-2=5\).

We need to make the variable, \(p\), the subject of the equation. \(2\) is being subtracted from \(p\), so we need to add \(2\) from both sides.

\[\begin{align*} p-2 & = 5\quad\textrm{add \(2\) to both sides}\\
p-2+2 & = 5+2\\
p & = 7
\end{align*}\]

The solution is \(p=7\).

Solve \(3g=18\).

We need to make the variable, \(g\), the subject of the equation. \(g\) is being multiplied by \(3\), so we need to divide by \(3\) on both sides.

\[\begin{align*} 3g & = 18\quad\textrm{divide both sides by \(3\)}\\
\frac{3g}{3} & = \frac{18}{3}\quad\textrm{cancel out \(3\) on left}\\
g & = 6
\end{align*}\]

The solution is \(g=6\).

Solve \(\dfrac{y}{4}=12\).

We need to make the variable, \(y\), the subject of the equation. \(y\) is being divided by \(4\), so we need to multiply by \(4\) on both sides.

\[\begin{align*} \frac{y}{4} = 12\quad\textrm{multiply both sides by \(4\)}\\
\frac{y}{4}\times4 & = 12\times4\quad\textrm{cancel out \(4\) on left}\\
y & = 48
\end{align*}\]

The solution is \(y=48\).

Exercise – solving basic linear equations

  1. Solve the following equations.
    1. \(x+3=7\)
    2. \(5-j=-2\)
    3. \(3c=12\)
    4. \(-r=-12\)
    5. \(\dfrac{m}{2}=-7\)
    6. \(-8u=12\)
    7. \(4g+4=16\)
    8. \(7-2w=1\)
    9. \(\dfrac{e}{2}-5=-8\)
    10. \(21-3t=12\)
    11. \(\dfrac{y}{5}-9=-5\)
    12. \(3-\dfrac{u}{2}=-7\)
  2. Solve the following equations.
    1. \(5i+2=i+10\)
    2. \(10p-11=2p-3\)
    3. \(5a-12=3a+6\)
    4. \(10d+10=0\)
    5. \(f+6=6f-9\)
    6. \(8-g=5g+14\)
    7. \(5h-2=7h-12\)
    8. \(6j+13=4j+13\)

    1. \(x=4\)
    2. \(j=7\)
    3. \(c=4\)
    4. \(r=12\)
    5. \(m=-14\)
    6. \(u=-\dfrac{3}{2}\)
    7. \(g=3\)
    8. \(w=3\)
    9. \(e=-6\)
    10. \(t=3\)
    11. \(y=20\)
    12. \(u=20\)
    1. \(i=2\)
    2. \(p=1\)
    3. \(a=9\)
    4. \(d=-1\)
    5. \(f=3\)
    6. \(g=-1\)
    7. \(h=5\)
    8. \(j=0\)

Solving more complex equations

With more complex equations, you will need to perform several operations. Remember the following:

  • If the variable appears in terms on both sides of the equation, you need to get these terms to one side of the equation and then solve for the variable.
  • If an equation involves brackets, the brackets should be expanded first.
  • If an equation involves fractions, you need to use a common denominator.

The strategy for solving equations is not part of the solution. Here, it is just shown to help you learn. As you become more experienced, you may even perform several operations in one step.

Example 1 – solving more complex equations

Solve \(2w-3=-17\).

We need to make the variable, \(w\), the subject of the equation. \(w\) is multiplied by \(2\), then \(3\) is subtracted. We need to work backwards to isolate \(w\).

\[\begin{align*} 2w-3 & = -17\quad\textrm{add \(3\) to both sides}\\
2w-3+3 & = -17+3\\
2w & = -14\quad\textrm{divide both sides by \(2\)}\\
\frac{2w}{2} & = \frac{-14}{2}\quad\textrm{cancel out \(2\) on left}\\
w & = -7
\end{align*}\]

The solution is \(w=-7\).

With more complex equations, it is a good idea to check your result. However, this is not necessary unless you are asked to check. In this example:

\[\begin{align*} \text{LHS} & =2w-3\\
& =2\times(-7)-3\\
& =-14-3\\
& =-17\\
& =\text{RHS}
\end{align*}\]

Our solution is correct.

Solve \(\dfrac{3d}{4}+5=7\).

We need to make the variable, \(d\), the subject of the equation. \(d\) is multiplied by \(3\), divided by \(4\), and then \(5\) is added. We need to work backwards to isolate \(d\).

\[\begin{align*} \frac{3d}{4}+5 & = 7\quad\textrm{subtract \(5\) from both sides}\\
\frac{3d}{4}+5-5 & = 7-5\\
\frac{3d}{4} & = 2\quad\textrm{multiply both sides by \(4\)}\\
\frac{3d\times4}{4} & = 2\times4\quad\textrm{cancel out \(4\) on left}\\
3d & = 8\quad\textrm{divide both sides by \(3\)}\\
\frac{3d}{3} & = \frac{8}{3}\quad\textrm{cancel out \(3\) on left}\\
d & = \frac{8}{3}
\end{align*}\]

The solution is \(d=\dfrac{8}{3}\).

We can check our result:

\[\begin{align*} \textrm{LHS} & = \frac{3}{4}\times\frac{8}{3}+5\\
& = 2+5\\
& = 7\\
& = \textrm{RHS}
\end{align*}\]

Solve \(3c+1=c-5\).

We need to make the variable, \(c\), the subject of the equation. \(c\) is on both sides, so we need to move any terms containing \(c\) to one side.

\[\begin{align*} 3c+1 & = c-5\quad\textrm{subtract \(c\) from both sides}\\
3c+1-c & = c-5-c\\
2c+1 & = -5\quad\textrm{subtract \(1\) from both sides}\\
2c+1-1 & = -5-1\\
2c & = -6\quad\textrm{divide both sides by \(2\)}\\
\frac{2c}{2} & = \frac{-6}{2}\quad\textrm{cancel out \(2\) on left}\\
c & = -3
\end{align*}\]

The solution is \(c=-3\).

We can check our result:

\[\begin{align*} \textrm{LHS} & =3c+1\\
& =3(-3)+1\\
& =-8\\
\textrm{RHS} & =c-5\\
& =(-3)-5\\
& =-8\\
& =\textrm{LHS}
\end{align*}\]

Solve \(3(5-2j)=33\).

We need to make the variable, \(j\), the subject of the equation. \(j\) is multiplied by \(2\), subtracted from \(5\), then multiplied by \(3\). We need to work backwards to isolate \(j\).

There are two ways to solve this one. One way is to expand the expression on the left to remove the brackets. Here, we will not show the calculation steps.

\[\begin{align*} 3(5-2j) & = 33\quad\textrm{expand the expression}\\
15-6j & = 33\quad\textrm{subtract \(15\) from both sides}\\
-6j & = 18\quad\textrm{divide both sides by \(-6\)}\\
j & = -3
\end{align*}\]

The other way is to get rid of the \(3\) outside of the brackets first by dividing both sides by \(3\).

\[\begin{align*} 3(5-2j) & = 33\quad\textrm{divide both sides by \(3\)}\\
5-2j & = 11\quad\textrm{subtract \(5\) from both sides}\\
-2j & = 6\quad\textrm{divide both sides by \(-2\)}\\
j & = -3
\end{align*}\]

Both strategies lead us to the same answer: \(j=-3\).

We can check our result:

\[\begin{align*} \textrm{LHS} & = 3(5-2j)\\
& = 15-6j\\
& = 15-6(-3)\\
& = 15+18\\
& = 33\\
& = \textrm{RHS}
\end{align*}\]

Solve \(2(3k-2)=5(k+7)\).

We need to make the variable, \(k\), the subject of the equation. \(k\) is on both sides, so we need to move any terms containing \(k\) to one side.

Like Example 4, there is more than one way to solve this equation. We will just show you one way here but have a think about other approaches.

\[\begin{align*} 2(3k-2) & = 5(k+7)\quad\textrm{expand the expressions}\\
6k-4 & = 5k+35\quad\textrm{subtract \(5k\) from both sides}\\
k-4 & = 35\quad\textrm{add \(4\) to both sides}\\
k & = 39
\end{align*}\]

The solution is \(k=39\).

We can check our result:

\[\begin{align*} \textrm{LHS} & =2(3k-2)\\
& = 6k-4\\
& = 6(39)-4\\
& = 230\\
\textrm{RHS} & = 5(k+7)\\
& = 5k+35\\
& = 5(39)+35\\
& = 230\\
& =\textrm{RHS}
\end{align*}\]

Solve \(\dfrac{h+1}{3}=\dfrac{h}{4}\).

We need to make the variable, \(h\), the subject of the equation. \(h\) is on both sides, so we need to move any terms containing \(h\) to one side.

\[\begin{align*} \frac{h+1}{3} & = \frac{h}{4}\quad\textrm{multiply each fraction to get the lowest common denominator (12)}\\
\frac{h+1}{3}\times\frac{4}{4} & = \frac{h}{4}\times\frac{3}{3}\\
\frac{4(h+1)}{12} & = \frac{3h}{12}\quad\textrm{multiply both sides by \(12\)}\\
4(h+1) & = 3h\quad\textrm{expand the expression}\\
4h+4 & = 3h\quad\textrm{subtract \(3h\) from both sides}\\
h+4 & = 0\quad\textrm{subtract \(4\) from both sides}\\
h & = -4
\end{align*}\]

The solution is \(h=-4\).

We can check our result:

\[\begin{align*} \textrm{LHS} & = \frac{h+1}{3}\\
& = \frac{(-4)+1}{3}\\
& = -1\\
\textrm{RHS} & = \frac{h}{4}\\
& = \frac{(-4)}{4}\\
& = -1\\
& = \textrm{LHS}
\end{align*}\]

Solve \(\dfrac{2z+11}{7}=\dfrac{z-3}{12}\).

We need to make the variable, \(z\), the subject of the equation. \(z\) is on both sides, so we need to move any terms containing \(z\) to one side.

\[\begin{align*} \frac{2z+11}{7} & = \frac{z-3}{12}\quad\textrm{multiply each fraction to get the lowest common denominator (84)}\\
\frac{2z+11}{7}\times\frac{12}{12} & = \frac{z-3}{12}\times\frac{7}{7}\\
\frac{12(2z+11)}{84} & = \frac{7(z-3)}{84}\quad\textrm{multiply both sides by \(84\)}\\
12(2z+11) & = 7(z-3)\quad\textrm{expand the expressions}\\
24z+132 & = 7z-21\quad\textrm{subtract \(7z\) from both sides}\\
17z+132 & = -21\quad\textrm{subtract \(132\) from both sides}\\
17z & = -153 \quad\textrm{divide both sides by \(17\)}\\
z & = -9
\end{align*}\]

The solution is \(z=-9\).

Solve \(\dfrac{3u}{4}-\dfrac{1}{3}=7\).

We need to make the variable, \(u\), the subject of the equation. \(u\) is on both sides, so we need to move any terms containing \(u\) to one side.

\[\begin{align*} \frac{3u}{4}-\frac{1}{3} & = 7\quad\textrm{multiply each fraction to get the lowest common denominator (12)}\\
(\frac{3u}{4}\times\frac{3}{3})-(\frac{1}{3}\times\frac{4}{4}) & = 7\times\frac{12}{12}\\
\frac{3u\times3}{12}-\frac{1\times4}{12} & = \frac{84}{12}\quad\textrm{multiply both sides by \(12\)}\\
9u-4 & = 84\quad\textrm{add \(4\) to both sides}\\
9u & = 88\quad\textrm{divide both sides by \(9\)}\\
u & = \frac{88}{9}
\end{align*}\]

Exercise – solving more complex linear equations

  1. Solve the following equations.
    1. \(3(2k-4)=18\)
    2. \(5(2z+9)=15\)
    3. \(3(x+4)=6\)
    4. \(3(c+3)+2(c-5)=4\)
    5. \(3(2v-3)+2(v-4)=-25\)
    6. \(3(b+4)=2(4b+1)\)
  2. Solve the following equations.
    1. \(\dfrac{9n}{5}-4=5\)
    2. \(\dfrac{4m}{3}-11=9\)
    3. \(1-\dfrac{9q}{2}=-8\)
    4. \(\dfrac{w-4}{2}=2\)
    5. \(\dfrac{3-2e}{11}=1\)
    6. \(\dfrac{3r+9}{5}=-3\)
    7. \(\dfrac{3t}{8}+4=1\)
    8. \(\dfrac{y}{3}=\dfrac{2}{9}\)
    9. \(\dfrac{5u-4}{4}=\dfrac{u-5}{5}\)
    10. \(\dfrac{2i+1}{7}=\dfrac{3i-5}{4}\)
    11. \(\dfrac{p+1}{3}+1=4\)
    12. \(2-\dfrac{5a-4}{4}=4\)
    13. \(\dfrac{d-3}{3}-4=\dfrac{d-2}{2}\)
    14. \(\dfrac{1-m}{5}-m=\dfrac{2m-1}{2}\)

    1. \(k=5\)
    2. \(z=-3\)
    3. \(x=-2\)
    4. \(c=1\)
    5. \(v=-1\)
    6. \(b=2\)
    1. \(n=5\)
    2. \(m=15\)
    3. \(q=2\)
    4. \(w=8\)
    5. \(e=-4\)
    6. \(r=-8\)
    7. \(t=-8\)
    8. \(y=\dfrac{2}{3}\)
    9. \(u=0\)
    10. \(i=3\)
    11. \(p=8\)
    12. \(a=-\dfrac{4}{5}\)
    13. \(d=-24\)
    14. \(m=\dfrac{7}{2}\)

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