Solving Simultaneous Equations Revision Notes for Grade 10 NSC Matric Mathematics

Solving Simultaneous Equations

What are simultaneous equations?

Simultaneous equations are a set of two or more equations that must be solved together to find the values of unknown variables. When you need to find two unknown variables, you need two independent equations that provide different information about those variables.

The key idea is that the solution consists of values for the unknown variables that satisfy both equations at the same time. For example, if we have variables $x$ and $y$ , we need to find specific values that work in both equations simultaneously.

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The word "simultaneous" means "at the same time" - this reminds us that our solution must work for all equations in the system together, not just one at a time.

A typical system of simultaneous equations looks like this:

$x + y = -1$
$3 = y - 2x$

In this case, we have two equations with two unknowns ( $x$ and $y$ ), and we need to find the specific values that make both equations true.

Method 1: Solving by substitution

The substitution method works by expressing one variable in terms of the other, then replacing it in the second equation. This transforms the problem into a single equation with one unknown.

Steps for substitution method:

Choose the simpler equation and rearrange it to express one variable in terms of the other
Replace this variable in the second equation
Solve the resulting equation (which now has only one unknown)
Substitute your answer back into the first equation to find the other variable
Check your solution by substituting both values into both original equations

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The substitution method is particularly useful when one of the equations is already solved for a variable, or can be easily rearranged to isolate a variable.

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Worked Example 1: Using substitution

Question: Solve for $x$ and $y$ :

$x - y = 1 \quad \text{...(1)}$

$3 = y - 2x \quad \text{...(2)}$

Solution:

Step 1: Use equation (1) to express $x$ in terms of $y$

$x = y + 1$

Step 2: Substitute this expression into equation (2)

$3 = y - 2(y + 1)$

$3 = y - 2y - 2$

$3 = -y - 2$

$5 = -y$

$y = -5$

Step 3: Substitute $y = -5$ back into equation (1)

$x = (-5) + 1 = -4$

Step 4: Check by substituting into both original equations

Equation (1): $-4 - (-5) = -4 + 5 = 1$ ✓
Equation (2): $3 = -5 - 2(-4) = -5 + 8 = 3$ ✓

Final answer: $x = -4$ , $y = -5$

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Worked Example 2: Another substitution problem

Question: Solve the system:

$4y + 3x = 100 \quad \text{...(1)}$

$4y - 19x = 12 \quad \text{...(2)}$

Solution:

Step 1: Express $x$ in terms of $y$ using equation (1)

$3x = 100 - 4y$

$x = \frac{100 - 4y}{3}$

Step 2: Substitute into equation (2)

$4y - 19\left(\frac{100 - 4y}{3}\right) = 12$

Multiply through by 3:

$12y - 19(100 - 4y) = 36$

$12y - 1900 + 76y = 36$

$88y = 1936$

$y = 22$

Step 3: Find $x$ by substituting back

$x = \frac{100 - 4(22)}{3} = \frac{100 - 88}{3} = \frac{12}{3} = 4$

Final answer: $x = 4$ , $y = 22$

Method 2: Solving by elimination

The elimination method involves making the coefficients of one variable the same in both equations, then adding or subtracting the equations to eliminate that variable.

Steps for elimination method:

Look at the coefficients of both variables in each equation
Choose which variable to eliminate (often the one with simpler coefficients)
Multiply one or both equations to make the coefficients of the chosen variable equal
Add or subtract the equations to eliminate that variable
Solve for the remaining variable
Substitute back to find the other variable

chatImportant

When using elimination, pay careful attention to the signs. If the coefficients of the variable you want to eliminate have the same sign, you subtract the equations. If they have opposite signs, you add the equations.

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Worked Example 3: Using elimination

Question: Solve the system:

$3x + y = 2 \quad \text{...(1)}$
$6x - y = 25 \quad \text{...(2)}$

Solution:

Step 1: Notice that the coefficients of $y$ are already $+1$ and $-1$ . We can eliminate $y$ by adding the equations:

\begin{array}{r} 3x + y = 2\\ 6x - y = 25\\ \hline 9x = 27 \end{array}

Step 2: Solve for $x$

$9x = 27$ $x = 3$

Step 3: Substitute back into equation (1)

$3(3) + y = 2$ $9 + y = 2$ $y = -7$

Final answer: $x = 3$ , $y = -7$

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Worked Example 4: Elimination with multiplication

Question: Solve the system:

$2a - 3b = 5 \quad \text{...(1)}$
$3a - 2b = 20 \quad \text{...(2)}$

Solution:

Step 1: Make the coefficients of $a$ equal by multiplying equation (1) by 3 and equation (2) by 2:

$6a - 9b = 15$
$6a - 4b = 40$

Step 2: Subtract the first from the second to eliminate $a$ :

$(6a - 4b) - (6a - 9b) = 40 - 15$
$-4b + 9b = 25$
$5b = 25$
$b = 5$

Step 3: Substitute back into original equation (1)

$2a - 3(5) = 5$
$2a - 15 = 5$
$2a = 20$
$a = 10$

Final answer: $a = 10$ , $b = 5$

Method 3: Solving graphically

Graphical solution involves plotting both equations as straight lines on the same coordinate system. The solution is the coordinates of the intersection point where the two lines meet.

Steps for graphical method:

Rearrange both equations into the form $y = mx + c$
Plot both lines on the same set of axes
Find the coordinates where the lines intersect
This intersection point gives you the solution $(x, y)$

Understanding the graphical approach

When you have two linear equations, each represents a straight line. The solution to the simultaneous equations is the point that lies on both lines - this is where they intersect.

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The graphical method gives you a visual understanding of simultaneous equations. If the lines are parallel (same gradient), there's no solution. If they're the same line, there are infinitely many solutions.

The graph shows two lines: $y = 2x - 3$ and $y = \frac{1}{2}x$ intersecting at the point $(2, 1)$ . This means $x = 2$ and $y = 1$ is the solution.

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Worked Example 5: Graphical solution

Question: Solve graphically:

$4y + 3x = 100 \quad \text{...(1)}$
$4y - 19x = 12 \quad \text{...(2)}$

Solution:

Step 1: Convert to $y = mx + c$ form

From equation (1): $4y = 100 - 3x$ , so $y = -\frac{3}{4}x + 25$

From equation (2): $4y = 12 + 19x$ , so $y = \frac{19}{4}x + 3$

Step 2: Plot both lines on the same axes

Step 3: The intersection point occurs at $(4, 22)$

Final answer: $x = 4$ , $y = 22$

You can verify this algebraically by substitution if needed. Both methods should give the same result!

Choosing the best method

Understanding when to use each method will help you work more efficiently:

Substitution works well when one equation can be easily rearranged to express one variable in terms of the other
Elimination is often faster when the coefficients are already similar or can be made equal with simple multiplication
Graphical method is useful for visualising the solution and checking your algebraic work

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In exams, you might be told which method to use. If not, choose the method that seems easiest based on the coefficients and structure of your equations.

Exam tips

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Essential exam strategies:

Always check your solution by substituting back into both original equations
Show your working clearly - marks are awarded for method even if the final answer is incorrect
When using elimination, be careful with signs when adding or subtracting equations
For graphical solutions, ensure your scales are appropriate and you can read the intersection point accurately

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Key Points to Remember:

Simultaneous equations require two equations to solve for two unknowns, and the solution must satisfy both equations at the same time
Substitution method: Express one variable in terms of the other, substitute, solve, then find the second variable
Elimination method: Make coefficients equal for one variable, then add or subtract equations to eliminate that variable
Graphical method: Plot both lines and find their intersection point - this gives you the solution coordinates
Always verify your answer by substituting back into both original equations to ensure both are satisfied

Solving Simultaneous Equations (Grade 10 NSC Matric Mathematics): Revision Notes