Systems of Linear Equations (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Solving an Equation with Brackets

Solve the equation $3(x - 2) + 5 = 2x + 7$.

Solution:

Write the equation:

$3(x - 2) + 5 = 2x + 7$

Expand the brackets:

$3x - 6 + 5 = 2x + 7$

Simplify the left-hand side:

$3x - 1 = 2x + 7$

Subtract $2x$ from both sides:

$x - 1 = 7$

Add $1$ to both sides:

$x = 8$

The solution is x = 8.

Worked Example: Solving an Equation with Fractions

Solve the equation $\frac{x+3}{4} - \frac{x-1}{2} = \frac{5}{2}$.

Solution:

Write the equation:

$\frac{x+3}{4} - \frac{x-1}{2} = \frac{5}{2}$

Multiply all terms by the LCD of $4$:

$(x + 3) - 2(x - 1) = 10$

Expand the brackets:

$x + 3 - 2x + 2 = 10$

Collect like terms:

$-x + 5 = 10$

Subtract $5$ from both sides:

$-x = 5$

Multiply both sides by $-1$:

$x = -5$

Checking your answer:

It's always good practice to verify your solution by substituting it back into the original equation.

Write the left-hand side:

$\text{LHS} = \frac{x+3}{4} - \frac{x-1}{2}$

Substitute $x = -5$:

$= \frac{(-5)+3}{4} - \frac{(-5)-1}{2}$

Simplify the numerators:

$= \frac{-2}{4} - \frac{-6}{2}$

Write with common denominator:

$= \frac{-2}{4} - \frac{-12}{4}$

Combine the fractions:

$= \frac{-2-(-12)}{4}$

Simplify:

$= \frac{10}{4} = \frac{5}{2}$

Compare with right-hand side:

$= \text{RHS}$

Since LHS = RHS, the solution is verified.

Worked Example: Solving by Substitution

Solve this system of equations using substitution:

$x - y = 5$ $2x + 3y = 6$

Solution:

Number the equations for reference:

Equation 1: $x - y = 5$

Equation 2: `$2x + 3y = 6$$

First, isolate $x$ in equation 1:

Write equation 1:

$x - y = 5$

Add $y$ to both sides:

$x = 5 + y$

Next, substitute $x = 5 + y$ into equation 2:

Write equation 2:

$2x + 3y = 6$

Substitute $x = 5 + y$:

$2(5 + y) + 3y = 6$

Expand the brackets:

$10 + 2y + 3y = 6$

Collect like terms:

$10 + 5y = 6$

Subtract $10$ from both sides:

$5y = -4$

Divide both sides by $5$:

$y = -\frac{4}{5}$

Finally, substitute $y = -\frac{4}{5}$ back into equation 1:

Write equation 1:

$x - y = 5$

Substitute $y = -\frac{4}{5}$:

$x - \left(-\frac{4}{5}\right) = 5$

Evaluate the adjacent signs:

$x + \frac{4}{5} = 5$

Subtract $\frac{4}{5}$ from both sides:

$x = \frac{21}{5}$

The solution is $\left(\frac{21}{5}, -\frac{4}{5}\right)$.

Verification:

Check the solution in both original equations.

Verifying equation 1: $x - y = 5$

$\text{LHS} = x - y$

$= \frac{21}{5} - \left(-\frac{4}{5}\right)$

$= \frac{21}{5} + \frac{4}{5}$

$= \frac{25}{5}$

$= 5$

$= \text{RHS}$

Verifying equation 2: $2x + 3y = 6$

$\text{LHS} = 2x + 3y$

$= 2 \times \frac{21}{5} + 3 \times \left(-\frac{4}{5}\right)$

$= \frac{42}{5} - \frac{12}{5}$

$= \frac{30}{5}$

$= 6$

$= \text{RHS}$

Since LHS = RHS for both equations, the solution is verified.

Worked Example: Basic Elimination

Solve the system using the elimination method:

$9x + y = 62$ $5x + y = 38$

Solution:

Number the equations:

Equation 1: $9x + y = 62$

Equation 2: $5x + y = 38$

The $y$-coefficients are already the same. Since both are positive, subtract equation 2 from equation 1 to eliminate the $y$-term:

$9x + y = 62$ $-(5x + y) = -(38)$

Line break after subtraction symbol:

$4x + 0y = 24$

Simplify:

$4x = 24$

Divide both sides by $4$:

$x = 6$

Now substitute $x = 6$ into equation 2:

Write equation 2:

$5x + y = 38$

Substitute $x = 6$:

$5(6) + y = 38$

Evaluate the multiplication:

$30 + y = 38$

Subtract $30$ from both sides:

$y = 8$

The solution is the ordered pair $(6, 8)$.

Worked Example: Elimination Requiring Multiplication

Solve the system using the elimination method:

$2x + 3y = 19$ $4x - y = 10$

Solution:

Number the equations:

Equation 1: $2x + 3y = 19$

Equation 2: $4x - y = 10$

There are two approaches to eliminate a variable. Since the first method is simpler, we'll use it:

Method 1: Multiply equation 1 by $2$ to make the $x$-coefficients both equal to $4x$, then subtract to eliminate $x$.

Write equation 1:

$2x + 3y = 19$

Multiply both sides by $2$:

$4x + 6y = 38$

The modified system is now:

Equation 1: $4x + 6y = 38$

Equation 2: $4x - y = 10$

Subtract equation 2 from equation 1:

$4x + 6y = 38$ $-(4x - y) = -(10)$

Line break:

$0x + 7y = 28$

Simplify:

$7y = 28$

Divide both sides by $7$:

$y = 4$

Now substitute $y = 4$ into equation 2:

Write equation 2:

$4x - y = 10$

Substitute $y = 4$:

$4x - 4 = 10$

Add $4$ to both sides:

$4x = 14$

Divide both sides by $4$:

$x = \frac{14}{4}$

Simplify:

$x = \frac{7}{2}$

The solution is $\left(\frac{7}{2}, 4\right)$.

Verification:

Verifying equation 1: $2x + 3y = 19$

$\text{LHS} = 2x + 3y$

$= 2 \times \frac{7}{2} + 3 \times 4$

$= 7 + 12$

$= 19$

$= \text{RHS}$

Verifying equation 2: $4x - y = 10$

$\text{LHS} = 4x - y$

$= 4 \times \frac{7}{2} - 4$

$= 14 - 4$

$= 10$

$$= \text{RHS}$`$

Since LHS = RHS for both equations, the solution is verified.