Simultaneous equations

Simultaneous equations

Simultaneous equations are used when we are dealing with more than one equation that all use the same unknowns. Usually we will have at least two unknowns in problems of this type. With the use of more equations we can work out the values of unknowns which would be impossible if we were to only be given one equation.

For example, if we were given the equation 5x+2y=120 the values of x and y could be a wide range of things and it would be impossible for us to solve with only this information. However, if we were then given a second equation, say 4x+2y=80 we could then find out the values of x and y exactly!

Solving using a common term

To be able to solve simultaneous equations we must put the two together but this can only be done if we have a common term. For example, if we have 4y-5x=100 and 7x-y=100 we can write that 4y-5x=7x-y since both of these sides equal 100 and so must be equal to each other. Therefore, we need to change one or both of the equations and rearrange so that they have a common term that is isolated to one side in both equations.

Example

Rewrite the following simultaneous equations as a single equation with only one unknown.

a) 3y+x=36 and 3y-4x=20

b) 6y+4x=20 and 5x-y=20

c) 6x-3y=21 and 3x-y=42

a) Here both of the equations have a term that is 3y so we can rearrange each separately to isolate this term. Doing this we get 3y=36-x and 3y=20+4x. Now we can combine the two simultaneous equations using 3y as the common term to get 36-x=20+4x which is then solved as a normal algebraic equation.

b) Here the shared term is 20 and the two equations are already arranged so that this term is isolated. So we simply put them together to get 6y+4x=5x-y.

c) In this pair of equations there are no obvious common terms, therefore we need to change one of the equations to make this so. By recognising we have 6x in one and 3x in another equation, we can simply multiply the second equation by 2 so that these terms are then equal. However, we must multiply the entire equation by 2 and not just one part. Doing this we get 6x-2y=84 instead of 3x-y=42. Now we can rearrange to isolate the common term of 6x giving the equations 6x=3y+21 and 6x-2y=84. Rearranging each gives 6x=3y+21 and 6x=2y+84. Combining these two then gives us 21+3y=84+2y.

Suppose we are solving a pair of simultaneous equations and when combining with a common term we are still left with an equation with two unknowns as in the example (b) above.

To solve this we must rearrange this new combined equation to get a link between the two unknowns and then substitute this into the original equations so that they only contain one unknown.

Example

How would you go about solving for x and y the simultaneous equations 6y+4x=20 and 5x-y+20?

Combining these two using 20 as the common term gives us 6y+4x=5x-y which we can change by rearranging and combining the terms to 7y=x. This new link between y and x can then be used by replacing the x value in one of the original equations with 7y. This gives us (using the first equation) 6y+4(7y)=20 which can then give us a value for y. Substituting this value back into one of the original equations would then give you the value of x.

Solving by addition and subtraction

Another way to solve a pair of simultaneous equations is to subtract or add them in the hope of eliminating one of the terms. To do this you must again have a common term between the two simultaneous equations, but this time we either subtract or add the two equations together.

To do this we must operate correctly by only operating algebraic terms with themselves. For example, we cannot operate 4x and 7y as these have different unknown values.

Example

Solve the simultaneous equations 12x+2y=210 and 8x+2y=140.

By subtracting these two equations we get:

Subtracting two equations

Clearly the two values of 2y cancel and subtracting other terms we get 4x=60 so we know that x=15. Using this we can work out by substituting the value of back into the original equations. Doing this we get 8(15)+2y=140 which tells us that y=10.

If we have a common term in both of the equations that have the same sign (i.e. they are both positive or both negative) then we must subtract the two equations. If we have common terms that have different signs (one is positive and the other negative) then we must add the two equations together. This makes sure that the common term is eliminated, leaving us with an equation that only has one unknown.

The same rules that we have already explained can still be used to make sure we have a common term. So we are allowed to multiply or divide one of the equations by any number as long as we do the same operation to every term.

Steps for solving simultaneous equations

1) Identify any common terms between the two equations. If there are none then change one of the equations (by multiplying or dividing everything in it by a number) so that there is. There will be many options of what to do here; try to choose the easiest manipulation that requires the least work. You may at times be required to multiply both of the equations by a number, therefore, when finding a common term you have to use your intuition as each problem will be different.

2) Look if the two common terms have the same sign in front of them. This should be done after aligning the equations so that all of the unknowns are on one side and the numbers on the other.

3) If the two common terms have the same sign, subtract the two equations. If the signs are different then add them.

4) You will be left with a new equation that has only one unknown; solve this by methods seen in the previous lesson.

5) Substitute the value you have worked out back into one of the original equations (it doesn’t matter which) and find the remaining value.

Now we will run through an example from start to finish so that you can be sure of the technique. The method we will use is the addition and subtraction method. This is probably the better of the two and should be used where possible.

Example

Solve the simultaneous equations 3x+7y=46 and x+9y=42

1) Firstly, there are no common terms between these two equations so we must change one or both of them. The easiest way to do this will be to multiply the second equation by 3 so that both equations have a term that is 3x, which is a much easier option than trying to get a common term from the y values. This gives us the equations 3x+7y=46 and 3x+27y=126.

2) The common term of 3x is positive in both of the equations, therefore we must subtract them. Which way around the subtraction is done will not affect the answer but we should look to avoid complications such as negatives in the answer.

3) Since the signs are the same we do:

Screenshot 2022 04 07 at 10.35.41

The equation we now have only has one unknown. By solving 20y=80 we get y=4

Now we just need to work out the value of x. To do this we must use the value we have for y and substitute it into one of the original equations and solve for x. This gives us 3x+7(4)=46 which, when we solve, we find 3x=18 so we know that x=6

Checking your answers

A quick check of your answer is always a good idea and is very quick for simultaneous equations. Simply take the values for x and y that you have found and substitute them into the two equations and make sure they make them true.

Inequalities

We have already looked at inequalities on this course in an earlier module but will recap what we already know here. Inequalities are used as a replacement for an equals sign and show whether something is more or less than something else. The four different inequalities are:

quicklatex.com 9680b67fb1196e6117b65a6e7bb7fff6 l3means ‘less than’

>” height=”13″ width=”12″ src=”https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-71d4d1558f2a68db2a080ec3845828b0_l3.png”>means ‘more than’</p><p><img loading=means ‘less than or equal to’

\geqmeans ‘more than or equal to’

When trying to find a solution to an inequality we use the same rules as when solving equations. Namely, that when we operate on one side we must do the exact same operation to the other side so that the inequality is still equivalent to the original. By doing this we can really just treat the inequality, no matter which of the four that it is, as an equals sign and solve for an unknown in exactly the same way as we would for a normal equation.

The only exception to this is obviously at the end when our value for an unknown will be more, less or equal to some number depending on the inequality sign in use.

Solving inequalities

As with solving equations, we must look to isolate the unknown terms in an inequality to solve. By doing this we will end up with a simple inequality where the unknown is more or less (or equal to) some value.

Example

Solve the inequality quicklatex.com ef268292807dab304f3a8d92605a083c l3 for the unknown x.

We must isolate the x term by moving the -8 to the other side of the inequality. This is done in exactly the same way as if we had an equals sign instead of <. So on the other side, -8 becomes positive and we get quicklatex.com fb00fec7bc9670d883a8ed50de497b20 l3 which is the same as quicklatex.com 382205b4c29f5b6a9a7f226643b239ef l3.

The exception to the rule

There are two exceptions to the rule where solving for inequalities which are not the same as with equations. Both require a little explanation but are fairly easy to remember.

When we multiply all terms in an inequality by a negative number, the inequality must be reversed. This means that if we multiply everything by a negative to reverse signs, we must change from ‘more than’ to ‘less than’ or vice versa.

Example

Solve quicklatex.com ef8a57f5fcc82975a2e4a42addfbfd75 l3

Multiplying everything by -1 would give us quicklatex.com c7facb5d251864d1a67d74a7e9f0bd7a l3 however, as stated above, we must also change the inequality sign when multiplying by a negative. So our answer is x>5″ height=”15″ width=”45″ src=”https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-a62009d1098ebd821e662b3774363606_l3.png”></p><p>When taking the reciprocal of an inequality we must also swap the inequality sign. (Remember that reciprocals are 1 divided by the original number).</p><h4 class=Example

Find an inequality for the reciprocal of quicklatex.com 1be8a3d369f83873300f9583b15bd704 l3

Since we need to find the reciprocal, we must put the values on the bottom of a fraction with 1 on the top. This gives us quicklatex.com d9cd4bd1dd011e611ff03bbe9f2fd502 l3 which is wrong! This is because we must swap the inequality sign also, so the correct answer is \frac{1}{x}>\frac{1}{4}” height=”23″ width=”45″ src=”https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-7be04bc2f305314beb29869bb7078127_l3.png”></p><h4 class=Answers

Even though we have used the same rules as dealing with equations, you must not forget that we are in fact solving inequalities. This means your answer should be in this form (e.g. quicklatex.com 16326126f1b541eb9babe9cdf8215696 l3). You will not be able to find the exact value of an unknown and must simply reduce the inequality to its simplest form.

Summary of inequalities

Here are the rules we must follow when solving inequalities for an unknown algebraic term:

Treat the inequality like an equals sign. With the only two exceptions being multiplying by a negative and reciprocals, we use the exact same rules when solving for inequalities as equations. This means whatever we do to one side of the inequality, we must do the same to the other.

Remember that the answer will be an inequality. This means that despite treating the sign like an equals, do not forget that it is in fact an inequality and your answer should reflect this.

If you must multiply by a negative or take the reciprocal, swap the inequality. This is the only rule that is different from solving equations.

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