Solving a system of simultaneous equations seems like a very daunting task at first. With more than one unknown quantity to find the value for, and apparently very little way of disentangling one variable from another, it can be a headache for people new to algebra. However, there are three different methods for finding the solution to the equation, with two depending more on algebra and being a bit more reliable, and the other turning the system into a series of lines on a graph.
Solve a system of simultaneous equations by substitution by first expressing one variable in terms of the other. Using these equations as an example:
x – y = 5
3_x_ + 2_y_ = 5
Re-arrange the simplest equation to work with and use this to insert into the second. In this case, adding y to both sides of the first equation gives:
x = y + 5
Use the expression for x in the second equation to produce an equation with a single variable. In the example, this makes the second equation:
3 × (y + 5) + 2_y_ = 5
3_y_ + 15 + 2_y_ = 5
Collect the like terms to get:
5_y_ + 15 = 5
Re-arrange and solve for y, starting by subtracting 15 from both sides:
5_y_ = 5 – 15 = −10
Dividing both sides by 5 gives:
y = −10 ÷ 5 = −2
So y = −2.
Insert this result into either equation to solve for the remaining variable. At the end of step 1, you found that:
x = y + 5
Use the value you found for y to get:
x = −2 + 5 = 3
So x = 3 and y = −2.
Check Your Answers
It’s good practice to always check that your answers make sense and work with the original equations. In this example, x – y = 5, and the result gives 3 – (−2) = 5, or 3 + 2 = 5, which is correct. The second equation states: 3_x_ + 2_y_ = 5, and the result gives 3 × 3 + 2 × (−2) = 9 – 4 = 5, which is again correct. If something doesn’t match up at this stage, you have made a mistake in your algebra.
Look at your equations to find a variable to remove:
x – y = 5
3_x_ + 2_y_ = 5
In the example, you can see that one equation has -y and the other has +2_y_. If you add twice the first equation to the second one, the y terms would cancel out and y would be eliminated. In other cases (e.g., if you wanted to eliminate x), you can also subtract a multiple of one equation from the other.
Multiply the first equation by two to prepare it for the elimination method:
2 × (x – y) = 2 × 5
So
2_x_ – 2_y_ = 10
Eliminate your chosen variable by adding or subtracting one equation from the other. In the example, add the new version of the first equation to the second equation to get:
3_x_ + 2_y_ + (2_x_ – 2_y_) = 5 + 10
3_x_ + 2_x_ + 2_y_ – 2_y_ = 15
So this means:
5_x_ = 15
Solve for the remaining variable. In the example, divide both sides by 5 to get:
x = 15 ÷ 5 = 3
As before.
Like in the previous approach, when you have one variable, you can insert this into either expression and re-arrange to find the second. Using the second equation:
3_x_ + 2_y_ = 5
So, since x = 3:
3 × 3 + 2_y_ = 5
9 + 2_y_ = 5
Subtract 9 from both sides to get:
2_y_ = 5 – 9 = −4
Finally, divide by two to get:
y = −4 ÷ 2 = −2
Solve systems of equations with minimal algebra by graphing each equation and looking for the x and y value where the lines intersect. Convert each equation to slope-intercept form (y = mx + b) first.
The first example equation is:
x – y = 5
This can be converted easily. Add y to both sides and then subtract 5 from both sides to get:
y = x – 5
Which has a slope of m = 1 and a y-intercept of b = −5.
The second equation is:
3_x_ + 2_y_ = 5
Subtract 3_x_ from both sides to get:
2_y_ = −3_x_ + 5
Then divide by 2 to get the slope-intercept form:
y = −3_x_/2 + 5/2
So this has a slope of m = -3/2 and a y-intercept of b = 5/2.
Use the y intercept values and the slopes to plot both lines on a graph. The first equation crosses the y axis at y = −5, and the y value increases by 1 every time the x value increases by 1. This makes the line easy to draw.
The second equation crosses the y axis at 5/2 = 2.5. It slopes downwards, and the y value decreases by 1.5 every time the x value increases by 1. You can calculate the y value for any point on the x axis using the equation if it’s easier.
Locate the point where the lines intersect. This gives you both the x and y coordinates of the solution to the system of equations.
