Simultaneous equations

Simultaneous equations

The term simultaneous equations refers to a set of two or more algebraic equations involving two or more variables. The solution to such a system is the set of values for the variables that simultaneously satisfy all the given equations.

Although many systems involve linear equations only, this section covers systems involving:

• Two linear equations in two variables
• One linear equation and one quadratic equation

Formulating and solving linear simultaneous equations

Real-life and mathematical problems can often be translated into algebraic form using simultaneous equations. Various techniques are used to solve them:

• Elimination Method
• Substitution Method
• Graphical Method
• Matrix Method (including Cramer's Rule)

In this section, we focus on the elimination and substitution methods.

Elimination method

The elimination method involves manipulating the equations so that one variable is eliminated when the equations are added or subtracted. This results in a single-variable equation, which can then be solved directly. Once one variable is found, it is substituted into either of the original equations to find the second variable.

Example 3.1:

Find two integers such that:

• Twice the first minus the second is 5
• Their sum is 1

Formulate the equations and solve them.

Solution:

Let the first integer be $x$ and the second be $y$. Then:

$$\begin{cases} 2x - y = 5 & \text{(1)} \\ x + y = 1 & \text{(2)} \end{cases}$$

To eliminate one variable, we can add or subtract the equations. First, let's eliminate $y$. Add equations (1) and (2):

Multiply equation (2) by 1 (no change) and equation (1) by 1:

$$\begin{align*} 2x - y &= 5 \quad \text{(from Eq. 1)} \\ x + y &= 1 \quad \text{(from Eq. 2)} \end{align*}$$

Add both equations:

$$(2x - y) + (x + y) = 5 + 1 \Rightarrow 3x = 6 \Rightarrow x = 2$$

Substitute $x = 2$ into equation (2):

$$x + y = 1 \Rightarrow 2 + y = 1 \Rightarrow y = -1$$

Therefore, the two integers are 2 and -1.

Substitution method

In the substitution method, one variable is first expressed in terms of the other using one of the equations. This expression is then substituted into the second equation to solve for one variable. Once that variable is found, it is substituted back to find the remaining variable.

Using Example 1 again:

Solve:

$$\begin{cases} 2x - y = 5 & \text{(1)} \\ x + y = 1 & \text{(2)} \end{cases}$$

using the substitution method.

Solution:

From equation (2):

$$x + y = 1 \Rightarrow y = 1 - x \quad \text{(3)}$$

Substitute (3) into equation (1):

$$2x - (1 - x) = 5 \Rightarrow 2x - 1 + x = 5 \Rightarrow 3x = 6 \Rightarrow x = 2$$

Substitute $x = 2$ into equation (3):

$$y = 1 - 2 = -1$$

Again, the two integers are 2 and -1.

Simultaneous equations involving linear and quadratic equations

Some problems involve solving a system of two equations in which one is linear and the other is quadratic. These can be solved using substitution or other algebraic methods.

Example 2

There are two numbers such that their sum is 10 and the sum of their squares is 52. Formulate simultaneous equations and find the two numbers.

Solution

Let the two numbers be $x$ and $y$. Then, the equations are:

$$\begin{aligned} x + y &= 10 \quad \text{(1)} \\ x^2 + y^2 &= 52 \quad \text{(2)} \end{aligned}$$

From equation (1), express $y$ in terms of $x$:

$$y = 10 - x \quad \text{(3)}$$

Substitute equation (3) into equation (2):

$$x^2 + (10 - x)^2 = 52$$

Expand and simplify:

$$x^2 + 100 - 20x + x^2 = 52$$ $$2x^2 - 20x + 100 = 52$$ $$2x^2 - 20x + 48 = 0$$ $$x^2 - 10x + 24 = 0$$

Factor the quadratic:

$$x^2 - 6x - 4x + 24 = 0$$ $$x(x - 6) - 4(x - 6) = 0$$ $$(x - 4)(x - 6) = 0$$

So, $x = 4$ or $x = 6$. Then from equation (3):

• If $x = 4$, then $y = 10 - 4 = 6$
• If $x = 6$, then $y = 10 - 6 = 4$

Therefore, the two numbers are 4 and 6.

Example 3

The sum of two whole numbers is 5, and the difference of their squares is 5. Find the two numbers.

Solution

Let the two numbers be $x$ and $y$, with $x > y$.

$$\begin{aligned} x + y &= 5 \quad \text{(1)} \\ x^2 - y^2 &= 5 \quad \text{(2)} \end{aligned}$$

From (1):

$$x = 5 - y \quad \text{(3)}$$

Substitute into (2):

$$(5 - y)^2 - y^2 = 5$$

Expand and simplify:

$$25 - 10y + y^2 - y^2 = 5$$ $$25 - 10y = 5$$ $$10y = 20 \Rightarrow y = 2$$

Then from (3):

$$x = 5 - 2 = 3$$

Therefore, the two numbers are 3 and 2.

Alternative method

Using the identity $x^2 - y^2 = (x - y)(x + y)$, equation (2) becomes:

$$(x - y)(x + y) = 5$$

Since $x + y = 5$, substitute:

$$(x - y)(5) = 5 \Rightarrow x - y = 1$$

Now solve the system:

$$\begin{aligned} x + y &= 5 \quad \text{(1)} \\ x - y &= 1 \quad \text{(4)} \end{aligned}$$

Add equations (1) and (4):

$$2x = 6 \Rightarrow x = 3$$

Substitute into (1):

$$3 + y = 5 \Rightarrow y = 2$$

Therefore, the two numbers are 3 and 2.

Example 4

The sum of two numbers is 12 and their product is 20. What is the difference between the two numbers?

Solution

Let the numbers be $a$ and $b$, with $a > b$. Then:

$$\begin{aligned} a + b &= 12 \quad \text{(1)} \\ ab &= 20 \quad \text{(2)} \end{aligned}$$

From (1):

$$a = 12 - b \quad \text{(3)}$$

Substitute into (2):

$$(12 - b)(b) = 20 \Rightarrow 12b - b^2 = 20 \Rightarrow b^2 - 12b + 20 = 0$$

Solve by factorization:

$$b^2 - 10b - 2b + 20 = 0 \Rightarrow b(b - 10) - 2(b - 10) = 0 \Rightarrow (b - 2)(b - 10) = 0$$

So $b = 2$ or $b = 10$. Then:

• If $b = 2$, then $a = 12 - 2 = 10$
• If $b = 10$, then $a = 12 - 10 = 2$

Since $a > b$, we take $a = 10$, $b = 2$.

The difference is: $a - b = 10 - 2 = 8$

Therefore, the difference between the two numbers is 8.