A system of equations—also called simultaneous equations or sets of equations—is just a fancy way of saying you’re working with multiple equations at once, looking for a common solution. Instead of solving one equation, you’re finding values that satisfy all the equations in the system. Pretty cool, right?

You’ll find systems of equations popping up in all sorts of real-world scenarios, from calculating budgets to figuring out rates and distances. Plus, they’re a staple on exams like the SAT, ACT, PSAT/NMSQT, GED, GRE, GMAT, AP Exams, PERT, Accuplacer, and even the MCAT. So, if you're aiming to ace those tests, having a solid grip on these equations will definitely give you an edge! Ready to master them? Let’s break it down step by step.

System of Equations

Algebraically, a system of equations consists of two or more linear equations that look for a common solution for the included variables. Some of the examples of system of equations are:

• 2x + y = 5, and -x + y = 2
• x + y = 6, and -3x + y = 2

Solutions to Systems of Equations

By solving a system of equations, it is meant that we find the value of the unknown variables from the given equations that satisfy both the equations. The value of the unknown variable is computed by keeping the equations on both the sides balanced.

The basic idea behind solving the equations is to find the value of variables satisfying all the equations simultaneously. There can be three different types of solutions namely:

• Unique solution
• Infinitely many solutions
• No solution

Unique Solution

By unique solution we mean that there exists one value for the variables that represent the point of intersection of the two lines formed by the equation. Also, by substituting the values of the solution in both the equations we get that the left hand and the right hand side of the equations are equal.

In case of linear equations in two variables, we get a common solution as a coordinate (x,y) satisfying both the equations.

Let the two equations be:

x + y = 6…(1)

-3x + y = 2…(2)

The two equations have a unique solution and intersect at a point on the graph. Let us plot these lines on a graph and check the intersection point.

For equation (1), we have:

And for equation (2), we have:

Let us plot these lines on a graph and check for a unique intersection point.

From the graph we can see that the point of intersection of the two lines is (1,5).

So, the lines give a unique solution.

Infinitely Many Solution

A system of equations is said to have infinitely many solutions if the solution set of the pair of lines has infinitely many points in it. Graphically we can say that the lines formed from the equation overlap or coincide with each other.

Let us understand this with an example:

2x – y = 4…(1)

6x – 3y = 12…(2)

The two equations have an infinitely many solutions and overlap each other on the graph. Let us plot these lines on a graph and check this.

For equation (1), we have:

And for equation (2), we have:

Let us plot these lines on graph and check for infinitely many solutions:

No Solution

A system of equations is said to have no solution with the lines plotted on the graph having no point of intersection. These are basically parallel lines that do not meet at any point.

Let us understand this using an example:

2x – y = 4…(1)

6x – 3y = 3…(2)

The two equations have a no solution and do not meet each other on the graph. Let us plot these lines on graph and check this.

For equation (1), we have:

And for equation (2), we have:

Let us plot these lines on graph and check for infinitely many solutions:

Solving System of Equations

Any system of equations can be easily solved by different methods. For solving linear equations in two variables we need at least two equations. In the same way for solving linear equations in three variables we need at least three such equations.

The three common methods to solve system of equations are:

• Substitution method
• Elimination method
• Graphical method

Let us discuss each method separately:

Substitution Method

In order to solve the equation using substitution method, following steps are followed:

• Find the value of one of the two variables from one of the equations in term of another variable.
• Substitute the value so obtained in the second equation, to get an equation in a single variable.
• Find the value of the unknown variable using the transposition method.
• Substitute the value so obtained in the first equation to obtain the value of the remaining variable.

Let us understand this using an example:

Example: Solve the pair of equations x + 2y = 3, and 7x – 15y = 2.

Solution: The given linear equations in two variables are:

x + 2y = 3…(1)

7x – 15y = 2…(2)

Considering equation (1), let us find the value of ‘x’ in terms of ‘y’, we get:

x = 3 – 2y…(3)

Substitute the value of x from (3) in (2), we get:

7(3-2y) – 15y = 2

21 – 14y – 15 y = 2

21 – 29y = 2

-29y = 2 – 21

-29y = -19

y = 19/29…(4)

Put this value of ‘y’ from (4) in (3)

x = 3 – 2(19/29)

x = 3 – 38/29

x = (87 – 38)/29

x = 49/29

So, the value of x = 49/29, and y = 19/29.

Elimination Method

In order to solve linear equations using elimination method, following steps are followed:

• Multiply both the equations with a constant except zero that makes coefficients of one of the variables in both the equations equal.
• Add or subtract the new equations so obtained to get the resultant in a single variable. And find the value of the variable using transposition.
• Put the value of the variable in the equations so obtained and find the value of the remaining variables.

Let us understand this using an example:

Example: Solve the pair of equations 2x + 5y = 20 and 3x + 6y = 12.

Solution: The given linear equations in two variables are:

2x + 5y = 20…(1)

3x + 6y = 12…(2)

In order to solve this pair of linear equations using the elimination method, we have to multiply equation (1) by 3 and equation (2) by 2. We get:

6x + 15y = 60…(3)

6x + 12y = 24…(4)

Subtract equation (4) from equation (3), we get:

3y = 36

y = 36/3 = 12

Put the value of y in equation (1), we get:

2x + 5(12) = 20

2x + 60 = 20

2x = -40

x = -20

So, the solution for the given pair of linear equations is x = -20 and y = 12.

Using Matrices

A pair of linear equations can be solved using matrices. In order to solve the equations using this method, equations are first expressed as standard form with the variables and constants on their respective sides.

Let the given equations be:

\(a_1x+b_1y+c_1z=d_1\)

\(a_2x+b_2y+c_2z=d_2\)

\(a_3x+b_3y+c_3z=d_3\)

This can be expressed as matrices like:

\(\left[\begin{matrix}a_1x+b_1y+c_1z\\a_2x+b_2y+c_2z\\a_3x+b_3y+c_3z\end{matrix}\right]=\left[\begin{matrix}d_1\\d_2\\d_3\end{matrix}\right]\)

\(\left[\begin{matrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{matrix}\right]+\left[\begin{matrix}x\\y\\z\end{matrix}\right]=\left[\begin{matrix}d_1\\d_2\\d_3\end{matrix}\right]\)

AX =B

Here,

A= \(\left[\begin{matrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{matrix}\right]\), X=\(\left[\begin{matrix}x\\y\\z\end{matrix}\right]\), and B = \(\left[\begin{matrix}d_1\\d_2\\d_3\end{matrix}\right]\)

So, \(X=A^{-1} B\)

Uses of System of Equations

The basic application of linear equations in real-life applications is where we have two variables out of which one is unknown. It can be used:

• To solve geometric problems
• Money problems
• Problems related to mixtures
• Distance rate and time problems
• Applications of linear equations in economics and business.

System of Equations Solved Examples

Que 1: In a \(\triangle ABC\), \(\angle C=3\angle B=2\left(\angle A+\angle B\right)\). Calculate the angles in \(\triangle ABC\).

Solution:

Let \(\angle A = x^{\circ} and \angle B = y^{\circ}\)

Then,

\(\angle C = 3\angle B = 3y^{\circ} \)

Now,

\(\angle A + \angle B + \angle C = 180^{\circ}\)

x + y + 3y = 180

x + 4y = 180…(1)

\(\angle C =2\left(\angle A+\angle B\right)\)

3y = 2(x + y)

3y = 2x + 2y

2x + 2y – 3y = 0

2x – y = 0…(2)

Solving equation (1) and (2) by substitution.

From equation (2) y = 2x

Put this value of y in equation (1):

x + 4(2x) = 180

x +8x = 180

9x = 180

x = 20

y = 2x = 2(20) = 40

So, \(\angle A = x = 20^{\circ}\)

\(\angle B = y = 40^{\circ}\)

\(\angle C = 3y = 3(40)=120^{\circ}\)

Que 2: Ramesh is three times as old as his son. Five years later he will be two and a half times as old as his son. Find the present age of Ramesh.

Solution 2: Let the present age of Ramesh be x years

And present age of his son is y years

Given that Ramesh is three times as old as his son.

So, x = 3y…(1)

Ramesh age after 5 years = (x + 5) years

Ramesh’s son age after 5 years = (y + 5) years

Given that five years later he will be two and a half times as old as his son.

So, x + 5 = 2.5 (y + 5)

x + 5 = 2.5 y + 12.5

x – 2.5y = 7.5…(2)

Add the value of x from equation (1) in equation (2), we get:

3y – 2.5y = 7.5

0.5y = 7.5

y = 7.5/0.5 = 15.

So, x = 3y = 3(15) = 45 years

So, Ramesh’s percentage is 45 years and his son is 15 years.

Conclusion

Systems of equations might seem tricky at first, but once you get the hang of them, they’re super useful—both in real life and on major exams like the SAT, ACT, and AP tests. Whether you’re solving for unique solutions, no solutions, or infinitely many, knowing methods like substitution, elimination, and graphical approaches will definitely give you an edge. Keep practicing, and soon solving these will feel like second nature!