What are simultaneous equations?
A set of two or more algebraic equations, each containing two or more common variables whose values satisfy all equations in the set, are referred to as simultaneous equations (also known as systems of equations). The number of variables in each given equation must be equal to or less than the number of equations in the set.
Below are some examples of simultaneous and non-simultaneous equations:
| Set of Equations | Remarks |
| 2x + 4y = 12 4x − 2y = 6 | Simultaneous equations. The number of variables and the number of equations is the same |
| 3a + 2b = 18 8a + 2c = 14 | Non- simultaneous equation. There are 3 variables, but there are only 2 equations. |
| 6h + 4i +3j = 8 4h + 7i +j = −2 2h – i = 3 | Simultaneous equations. The number of variables and the number of equations are the same. The number of variables in each equation must be equal to or less than the number of equations in the set. |
| 2m – 3n= 2 4p+ 2r = 3 3m-2p+5r=7 | Non- simultaneous equation. There are 4 variables, but there are only 3 equations. |
How to solve simultaneous equations graphically?
Knowing how to plot equation graphs is essential in solving simultaneous equations graphically. To find solutions by graphing, we need to graph the equations and look for the point where the graphs intersect one another. The point of intersection is the solution point.
To solve simultaneous equations graphically, you may follow these steps:
Step 1: Graph the given equations in a single coordinate plane. You may use any appropriate method for graphing.
Step 2: Look for the point where all the lines intersect. The point of intersection is the solution point to the given simultaneous equations. If the lines do not intersect, then the simultaneous equations have no solution. If the lines are coincident or above each other, then the simultaneous equations have infinitely many solutions.
Step 3: Determine the value of x and y from the coordinates of the point of intersection.
Step 4: Check if the solution you got satisfies both equations by plugging in the x and y that you got to the original equations.
Example #1
Find the values of x and y in the equations x + y = 12 and x = 2y −3.
Solution
Step 1: Graph x + y = 12 by finding its x- intercept and y- intercept.
To find the x- intercept, let y= 0.
x + y = 12
x + 0 = 12
x = 12
Hence, the x-intercept is at (12, 0)
To find the y- intercept, let x= 0.
x + y = 12
0 + y = 12
y = 12
Hence, the y-intercept is at (0, 12)
Plot the 2 points and draw a line to connect them. This is the graph x + y = 12.
Step 2: Graph x = 2y −3 by finding its x- intercept and y- intercept.
To find the x- intercept, let y= 0.
x = 2y −3
x = 2(0) −3
x = −3
Hence, the x-intercept is at (−3, 0)
To find the y- intercept, let x= 0.
x = 2y −3
0 = 2y −3
2y = 3
y = $\frac{3}{2}$
Hence, the y-intercept is at (0, $\frac{3}{2}$)
Plot the 2 points and draw a line to connect them. This is the graph of x = 2y −3.
Step 3: Look for the point of intersection of the two lines. The lines intersect at (7, 5). Hence we can say that x= 7 and y = 5 is the solution to the given equations.
Step 4: Check if the solution you got satisfies both equations by plugging in the values of x and y that you got to the original equations.
x + y = 12
(7) + (5) = 12
12 = 12 ✔
x = 2y −3
(7) = 2(5) −3
7 = 10 −3
7 = 7 ✔
Therefore, the solution to the simultaneous equations is x = 7 and y = 5.
Example #2
Find the values of x and y in the equations 2x + 4y = 6 and 3x + 2y = 13.
Solution
Step 1: Graph 2x + 4y = 6 by finding its x- intercept and y- intercept.
To find the x- intercept, let y= 0.
2x + 4y = 6
2x + 4(0) = 6
2x = 6
x = 3
Hence, the x-intercept is at (3, 0)
To find the y-intercept, let x= 0.
2x + 4y = 6
2(0) + 4y = 6
4y = 6
y = 3/2
Hence, the y -intercept is at (0, $\frac{3}{2}$)
Plot the 2 points and draw a line to connect them. This is the graph 2x + 4y = 6.
Step 2: Graph 3x + 2y = 13 by finding its x- intercept and y- intercept.
To find the x- intercept, let y= 0.
3x + 2y = 13
3x + 2(0) = 13
3x = 13
x = $\frac{13}{3}$
Hence, the x-intercept is at ($\frac{13}{3}$, 0)
To find the y-intercept, let x= 0.
3x + 2y = 13
3(0) + 2y = 13
2y = 13
y = $\frac{13}{2}$
Hence, the y-intercept is at (0, $\frac{13}{2}$)
Plot the 2 points and draw a line to connect them. This is the graph of 3x + 2y = 13.
Step 3: Look for the point of intersection of the two lines. The lines intersect at (5, −1). Hence, we can say that x = 5 and y = −1 is the solution to the given equations.
Step 4: Check if the solution you got satisfies both equations by plugging in the values of x and y that you got to the original equations.
2x + 4y = 6
2(5) + 4(−1) = 6
10 − 4 = 6
6 = 6 ✔
3x + 2y = 13
3(5) + 2(−1) = 13
15 − 2 = 13
13 = 13 ✔
Therefore, the solution to the simultaneous equations is x = 5 and y = −1.
Note that there are other methods in graphing equations in case that getting the intercepts and plotting them is burdensome.
How to solve simultaneous equations algebraically?
The use of algebra in solving simultaneous equations enables us to obtain the exact solution to the given equations. There are two known algebraic methods to solve simultaneous equations, the Substitution Method and the Elimination Method. The same examples will be used for you to see that the solution to the simultaneous equations discussed above is unique.
Substitution Method
To solve the simultaneous equations by substitution method, we must find the value of one variable in terms of the other, say find y in terms of x (or vice versa), for one of the given equations and then substitute the value you got into the other equation to solve for the remaining variable or unknown value.
Below are the steps in solving simultaneous equations by substitution.
- Use one of the equations to find x in terms of y or y in terms of x.
- Substitute this in the other equation to get an equation in terms of the single variable.
- Solve for the value of the variable.
- Substitute the value of the variable to any of the equations to get the value of another variable.
Example 1
Find the values of x and y in the equations x + y = 12 and x = 2y −3.
Solution
| Substitution Method Process | Step-by-step explanation |
| x + y = 12 → ① x = 2y −3 → ② | Label the equations as shown. Solve one of the given equations for one variable. In this case, we already have the value of x which is 2y – 3. |
| (2y − 3) + y = 12 3y − 3 = 12 → ③ | Substitute the value of x in ② to the x in ①. The result will now be our ③. |
| 3y − 3 = 12 → add 3 to both sides of the equation 3y − 3 + 3 = 12 + 3 3y = 12 + 3 3y = 15 $\frac{3y}{3} = \frac{15}{3}$ → divide the equation by 3 y = 5 | Solve for the value of y in ③ using algebraic processes. |
| x = 2(5) −3x = 10 −3 x = 7 | Substitute the value of y from ③ into the y in either of the original equations. In this case, we substitute y = 5 to ② to find the value of x. |
| x + y = 12 → ① (7) + (5) = 1212 = 12 ✔ x = 2y −3 → ② (7) = 2(5) −37 = 10 −3 7 = 7 ✔ | Plugin the values of x and y that you got to the original equations to check if they satisfy both equations. If yes, then they are the solution to the given simultaneous equation. |
| Therefore, x = 7 and y = 5. |
Example 2
Find the values of x and y in the equations 2x + 4y = 6 and 3x + 2y = 13.
Solution
| Substitution Method Process | Step-by-step explanation |
| 2x + 4y = 6 → ① 3x + 2y = 13 → ② | Label the equations as shown. |
| 2x + 4y = 6 → divide the equation by 2 2 2 2 x + 2y = 3 → subtract 2y to both sides of the equation x + 2y − 2y = 3 − 2y x = 3 − 2y → ① | Solve one of the given equations for one variable. In this case, we chose to manipulate ① using algebraic processes to find the value of x in terms of y. |
| 3(3 − 2y) + 2y = 13 9 − 6y + 2y = 13 9 − 4y = 13 → ③ | Substitute the value of x in ① to the x in ②. The result will now be our ③. |
| 9 − 4y = 13→ subtract 9 to both sides of the equation 9 − 4y – 9 = 13 − 9 $\frac{− 4y}{-4} = \frac{4}{-4}$ → divide the equation by 4 y = −1 | Solve for the value of y in ③ using algebraic processes. |
| 2x + 4(−1) = 6 2x – 4 = 6 → add 4 to both sides of the equation 2x – 4 + 4 = 6 + 4 2x = 6 + 4 2x = 10 $\frac{2x}{2} = \frac{10}{2}$ → divide the equation by 2 x = 5 | Substitute the value of y from ③ into the y in either of the original equations. In this case, we substitute y = −1 to ① to find the value of x. |
| 2x + 4y = 6 → ① 2(5) + 4(−1) = 6 10 − 4 = 6 6 = 6 ✔ 3x + 2y = 13 → ② 3(5) + 2(−1) = 13 15 − 2 = 13 13 = 13 ✔ | Plugin the values of x and y that you got to the original equations to check if they satisfy both equations. If yes, then they are the solution to the given simultaneous equation. |
| Therefore, x = 5 and y = − 1. |
Elimination Method
To solve simultaneous equations by the elimination method, you must make the coefficients of one of the variables the same in both equations. Then you either add or subtract the equations, whichever is appropriate, to form a new equation with one variable. The other variable was eliminated, thus the term elimination. Consequently, substitute the value found for the remaining variable in one of the given equations and find the other variable’s value.
Below are the steps in solving simultaneous equations by elimination.
- Arrange the equations in form, ax+by=c.
- Check if adding or subtracting the equations will result in the cancellation of a certain variable.
- If step 2 does not work, multiply one or both equations by the coefficient of either x or y.
- Solve for the value of the variable.
- Substitute the value of the variable to any of the equations to get the value of another variable.
Example #1
Find the values of x and y in the equations x + y = 12 and x = 2y −3
Solution
| Elimination Method Process | Step-by-step explanation |
| x + y = 12 → ① x = 2y −3 → ② | Label the equations as shown. |
| x = 2y −3 x − 2y = 2y −3 − 2y x − 2y = −3 → ② | Make sure that both equations are of the same form. In this case, we chose to manipulate ② using algebraic processes to convert it to the form of ①. |
| x + y = 12x − 2y = −3 | Set up both equations vertically such that the like terms are arranged column-wise. |
| x + y = 12 −1(x − 2y = −3) → multiply by –1 ———————— x + y = 12 − x + 2y = +3 ———————— 0 + 3y = 15 → ③ $\frac{3y}{3} = \frac{15}{3}$ → divide both sides by 3 y = 5 | Manipulate the equations algebraically such that when we add or subtract the two equations, one of the variables will be eliminated. In this case, we multiply ② by – 1 and then add the equations to eliminate x and solve for the value of y. |
| x + (5) = 12 x + 5 = 12→subtract 5 to both sides of the equation x + 5 − 5 = 12 – 5 x = 7 | Substitute the value of y from ③ into the y in either of the original equations. In this case, we substitute y = 5 to ① to find the value of x. |
| x + y = 12 → ① (7) + (5) = 12 12 = 12 ✔ x = 2y −3 → ② (7) = 2(5) −3 7 = 10 −3 7 = 7 ✔ | Plugin the values of x and y that you got to the original equations to check if they satisfy both equations. If yes, then they are the solution to the given simultaneous equation. |
| Therefore, x = 7 and y = 5. |
Example #2
Find the values of x and y in the equations 2x + 4y = 6 and 3x + 2y = 13.
Solution
| Elimination Method Process | Step-by-step explanation |
| 2x + 4y = 6 → ① 3x + 2y = 13 → ② | Label the equations as shown. Make sure that both equations are of the same form. In this case, they already are. |
| 2x + 4y = 6 3x + 2y = 13 | Set up both equations vertically such that the like terms are arranged column-wise. |
| 2x + 4y = 6 3x + 2y = 13 → multiply by –2 ————– 2x + 4y = 6 –2(3x + 2y = 13) ——————- 2x + 4y = 6 – 6x – 4y = – 26 ——————- – 4x – 0 = – 20 → ③ $\frac{– 4x}{-4} = \frac{– 20}{-4}$ → divide both sides by –4 x = 5 | Manipulate the equations algebraically such that when we add or subtract the two equations, one of the variables will be eliminated. In this case, we multiply ② by –2 and then add the equations to eliminate y and solve for the value of x. |
| 2(5) + 4y = 6 10 + 4y = 6→subtract 10 to both sides of the equation 10 + 4y – 10 = 6 – 10 $\frac{4y}{4} = \frac{– 4}{4}$ → divide both sides by 4 y = – 1 | Substitute the value of x from ③ into the x in either of the original equations. In this case, we substitute x = 5 to ① to find the value of y. |
| 2x + 4y = 6 → ① 2(5) + 4(−1) = 6 10 − 4 = 6 6 = 6 ✔ 3x + 2y = 13 → ② 3(5) + 2(−1) = 13 15 − 2 = 13 13 = 13 ✔ | Plugin the values of x and y that you got to the original equations to check if they satisfy both equations. If yes, then they are the solution to the given simultaneous equation. |
| Therefore, x = 5 and y = − 1. |
Common mistakes when solving simultaneous equations
Below are some common mistakes that you should watch for when solving simultaneous equations.
- Incorrect elimination of a variable.
The use of subtraction to eliminate one variable when you should be adding and vice-versa.
- Errors in operating negative numbers.
Making small mistakes when performing mathematical operations with negative numbers can lead to an erroneous answer. Work out the calculation separately so you can minimize errors.
- Not multiplying every term in the equation.
When multiplying a number to an equation, make sure to multiply it to every term on both sides of the equation.
- Not checking the answer using substitution.
Make it a habit to check and verify your solution if it satisfies all of the given equations by substituting the value of the variables you got to the original equations. This can help you quickly spot any existing errors.
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