SIMULTANEOUS EQUATIONS
Simultaneous equations are two algebraic equations that share common variables (such as x and y). They are called simultaneous equations because the equations are solved at the same time.
The following is an example of simultaneous equations: 2h+4k=4 5h+8k=3 At first, this might seem quite confusing. If we have 2 different variables which we don’t know, then how could we use them to solve 2 equations? We will answer this question in this topic.
Though they may appear simple, simultaneous equations find applications in a diverse array of fields. Whether it’s designing bridges, predicting market behavior, or understanding the forces of nature in physics, simultaneous equations play a crucial role in shaping our understanding and solving practical challenges.
There are 3 main ways of solving simultaneous equations at the Form 1 level:
- Elimination method
- Substitution method
- Graphical method
We will learn a fourth way in Form 3, but for now let us focus on these three.
Elimination Method
To solve simultaneous equations by the elimination method, we eliminate a variable, find the value of the other variable then use this value to find the value of the first variable.
We follow the following procedure:
- Add or subtract the equations to get rid of one of the variables.
- Find the value of one variable.
- Find the value of the remaining variables using substitution.
- Clearly state the final answer.
- Check your answer by substituting both values into either of the original equations.
Let us try out some examples.
Example 1
Solve the following simultaneous equations using elimination method.
2u-w=11 3u+w=49
The first thing we need to do is eliminate one of the variables by adding or subtracting the equations from each other. Here, it looks like it would be easier to eliminate w, because it has the same coefficient in both equations.
To eliminate w, we need to add the two equations to each other. \begin{align*}2u-w&=11\\+\quad 3u+w&=49\\ \hline 5u\qquad &=60\end{align*}
We can now solve for u. 5u=60 u=12
We can substitute the value of u into one of the original equations to solve for w: 2u-w=11 2(12)-w=11 24-w=11 w=24-11 w=13
And there we have it! Our final answer is \boxed{u=12} and \boxed{w=13} . The values u=12 and w=13 are said to satisfy the pair of simultaneous equations. Try substituting these values into the original equations to confirm the answer.
Example 2
Solve using elimination method:
3x+4y=18 5x+6y=28
Our first step is to eliminate one of the variables. To eliminate a variable, its coefficient in both equations must be the same. But we seem to have a problem here. All the variables have different coefficients. A question arises: how can we manipulate the equations to make one of the variables have the same coefficient in both equations?
Let us eliminate the y variable. (We are choosing to eliminate y just for demonstration, you can also eliminate x just as easily.) To make the variable have the same coefficient, we can multiply the first equation by 3, and the second equation by 2.
We find the numbers 3 and 2 by looking at the LCM of the coefficients of y. The LCM of 4 and 6 is 12.
Next, we need to find a certain number we can multiply by each equation to make the coefficient of y become 12 in both equations. In the first equation, to get from 4 to 12, we need to multiply by 3. Similarly, to get from 6 to 12 in the second equation, we must multiply by 2.
\begin{align*}3(3x+4y&=18) &\qquad\qquad\qquad 2(5x+6y&=28) \\ 9x+12y&=54 &\qquad\qquad\qquad 10x+12y&=56\end{align*}Now, to eliminate the y variable, we subtract one equation from the other. Here, we can subtract the first equation from the second one just so we don’t have to deal with negative values.\begin{align*}10x+12y&=56\\ -\quad 9x+12y&=54\\ \hline x\qquad\quad&=2\end{align*}
Now that we have x=2, we can substitute this value into one of the original equations to find y.
3x+4y=183(2)+4y=186+4y=184y=12y=3
Thus, the values that satisfy this pair of simultaneous equations are \boxed{x=2} and \boxed{y=3} .
Substitution Method
In this method, we express one variable in terms of another variable, then substitute the expression into the other equation. Let us look at an example:
Example
Solve the following simultaneous equations using the substitution method.
2m+4n=123m-2n=10
The first step is expressing one variable in terms of another variable. Let us try expressing n in terms of m. From the second equation: 3m-2n=102n=3m-10n=\frac{3m-10}{2}
We can now substitute this value of n into the first equation:2m+4n=122m+4\left(\frac{3m-10}{2}\right)=122m+6m-20=128m=32m=4
Now that we have the value of m, we can substitute it back into the expression for n: n=\frac{3m-10}{2}n=\frac{3(4)-10}{2}n=\frac{12-10}{2}n=\frac{2}{2}n=1 Our final answer is therefore \boxed{m=4} and \boxed{n=1} .
Graphical Method
You may have noticed that each equation in a simultaneous equation is the equation of a straight line. Because of this, we can solve simultaneous equations by graphing them and noting the point at which they intersect.
Example
Using the grid provided, solve the simultaneous equations.
x-4y=-5 -x+2y=1
To solve this pair of simultaneous equations using graphical methods, we must first express each equation in the form of the equation of a straight line. x-4y=-54y=x+5y=\frac{1}{4}x+\frac{5}{4}
To graph this line, we can create a table of x and y values that satisfy this equation. These values are found by plugging different values of x into the equation. We find 3 sets of x and y values to draw the line. This is because if we only found 2 and miscalculated one of them, the line would be wrong and we would not be able to tell. By finding 3, if we miscalculated one point, it would clearly not lie on the line, signalling a mistake.
Let us now plot this line on a graph.
We then repeat this process for the other line. -x+2y=1 2y=x+1 y=\frac{1}{2}x+\frac{1}{2}
Plotting this line on the same axes:
We see that the two lines intersect at the point (3,2). Thus, the solution to our pair of simultaneous equations is \boxed{x=3} and \boxed{y=2} .
Let us look at some more practical applications of simultaneous equations.
Example 1
Gela had two bags A and B, containing sugar. If he removed 2kg of sugar from bag A and added it to bag B, the mass of sugar in bag B would be four times the mass of sugar in bag A. If he added 10kg of sugar to the original amount of sugar in each bag, the mass of sugar in bag B would be twice the mass of sugar in bag A. Calculate the original mass of sugar in each bag.
We can let the original mass of bag A be A\text{ kg} and the original mass of bag B be B\text{ kg}.
In the question, we are told that removing 2 kg from bag A and moving it to B would make B 4 times heavier than A. This can be represented algebraically as: B+2=4(A-2)B+2=4A-8B-4A=-10
We are also told that adding 10 kg to each original amount would make bag B 2 times heavier than A. We can represent this as follows: B+10=2(A+10)B+10=2A+20B-2A=10
And just like that, we have formed a pair of simultaneous equations! We can solve them as usual, using either the substitution or elimination method.
\begin{align*}B-2A&=10\\ -\quad B-4A&=-10 \\ \hline 2A&=20 \end{align*}A=10
We can now substitute this value of A into one of the equations to find B: B-2A=10B-2(10)=10B-20=10B=30 Our final answer is \boxed{A=10\ kg} and \boxed{B=30\ kg} .
Example 2
A trader bought two types of bulbs R and D at Ksh 60 and Ksh 56 respectively. She bought a total of 50 bulbs of both types for Ksh 2872. Determine the number of type R bulbs that she bought.
This example involves a slightly different technique. While it may initially look like it would need simultaneous equations, it actually requires you to express both using the same variable. Let us see how this is done.
We can let the number of type R bulbs bought be x. Since the trader bought a total of 50 bulbs, the number of type D bulbs would be 50-x. The total amount of money spent on type R bulbs is \text{Ksh }60x, while that spent on type D bulbs is \text{Ksh }56(50-x). We can form a new equation for the total amount of money spent using these expressions: 60x+56(50-x)=2872
This is an equation with just one variable, so we can easily solve for x. 60x+2800-56x=28724x=72 x=18 Thus, the number of type R bulbs is 18.