Solving Linear and Quadratic Simultaneous Equations

Setting up Algebraic Equations

To solve a mathematical problem, you must often first translate a real world situation or a procedure into an algebraic expression or formula. In solving an equation, the goal is to isolate the unknown quantity to find its specific value. This is typically achieved by performing the same operation, such as adding, subtracting, multiplying, or dividing, to both sides of the equation to maintain balance while simplifying the terms.

Example: Setting up a linear equation

In a shop, the cost of one pen is 4 times the change received from $£15$ after buying one pen. What is the cost of one pen?

1. Identify the unknown: Let $x$ be the cost of one pen in pounds ($£$).
2. Express the change: The change from $£15$ is $(15 - x)$.
3. Form the equation: $x = 4(15 - x)$.
4. Solve by multiplying out the bracket: $x = 60 - 4x$.
5. Add $4x$ to both sides: $5x = 60$.
6. Divide both sides by 5: $x = 12$.

The cost of one pen is $£12$.

Graphical Solution of Simultaneous Linear Equations

Simultaneous equations can be solved graphically by plotting each equation as a line on a coordinate grid. The solution is the point (or points) where the lines intersect, as this coordinate satisfies both equations at once.

Example: Graphical solution of linear equations

Solve the following pair of equations graphically: $3x + 2y = 9$ and $x - y = -2$.

First, find coordinates for each line:

• For $3x + 2y = 9$: When $y = 0, x = 3$. When $x = 0, y = 4.5$.
• For $x - y = -2$: When $x = 0, y = 2$. When $y = 0, x = -2$.

Plotting these points and drawing the lines shows they intersect at a single point.

From the graph, the intersection point is $(1, 3)$, so the solution is $x = 1$ and $y = 3$.

Algebraic Solution of Simultaneous Linear Equations

To solve linear simultaneous equations in two unknowns algebraically, you must eliminate one of the variables. This can be done through substitution or by creating a linear combination of the two equations.

Example: Algebraic solution by substitution

Solve: $3x - 4y = 7$ (Equation i) and $x + 2y = 9$ (Equation ii).

1. Rearrange (ii) to express $x$ in terms of $y$: $x = 9 - 2y$ (Equation iii).
2. Substitute (iii) into (i): $3(9 - 2y) - 4y = 7$.
3. Expand the brackets: $27 - 6y - 4y = 7$.
4. Simplify: $27 - 10y = 7$.
5. Subtract 27 from both sides: $-10y = -20$.
6. Divide by $-10$: $y = 2$.
7. Substitute $y = 2$ back into (iii) to find $x$: $x = 9 - 2(2) = 5$.

Check by substituting both into (i): $3(5) - 4(2) = 15 - 8 = 7$, which is correct. The solution is $x = 5, y = 2$.

Algebraic Solution of One Linear and One Quadratic Equation

When one equation is linear and the other is quadratic, use the method of substitution. Rearrange the linear equation to isolate one variable, then substitute this into the quadratic equation. This usually results in a new quadratic equation in one variable, which typically yields two pairs of solutions.

Example: Solving a linear-quadratic system

Solve: $x + 4y = 7$ (Equation i) and $x^2 + 4xy + 2y^2 = 1$ (Equation ii).

1. Rearrange (i) to isolate $x$: $x = 7 - 4y$ (Equation iii).
2. Substitute (iii) into (ii): $(7 - 4y)^2 + 4(7 - 4y)y + 2y^2 = 1$.
3. Expand the terms: $(49 - 56y + 16y^2) + (28y - 16y^2) + 2y^2 = 1$.
4. Collect like terms: $2y^2 - 28y + 49 = 1$.
5. Rearrange into standard quadratic form: $2y^2 - 28y + 48 = 0$.
6. Divide by 2: $y^2 - 14y + 24 = 0$.
7. Factorise: $(y - 12)(y - 2) = 0$, so $y = 12$ or $y = 2$.
8. Find corresponding $x$ values using (iii):
   • If $y = 2, x = 7 - 4(2) = -1$.
   • If $y = 12, x = 7 - 4(12) = -41$.

The solutions are $x = -1, y = 2$ and $x = -41, y = 12$.

Graphical Solution of Non-Linear Systems

Approximate solutions for systems involving a quadratic equation can be found by plotting both the line and the curve. The coordinates of the intersection points provide the solutions.

Example: Graphical linear and quadratic solution

Find all approximate solutions for $x - y = -2$ and $y = x^2 + 4x + 1$ in the range $-5 \leq x \leq 3$.

By plotting the points for both equations within the specified range:

The intersections occur at approximately $x = -3.3, y = -1.3$ and $x = 0.3, y = 2.3$.

Modelling Real Situations

Worded problems can be converted into simultaneous equations to determine unknown costs or quantities.

Example: Pricing paint shades

Two shades of purple are made of red and blue paint:

• 10 litres of Imperial Purple (7 litres blue, 3 litres red) costs $£16.50$.
• 10 litres of Royal Purple (5 litres blue, 5 litres red) costs $£17.50$. How much does 10 litres of red paint cost?

1. Define variables: Let $x$ be the cost of 1 litre of blue and $y$ be the cost of 1 litre of red.
2. Set up equations:
   • $7x + 3y = 16.50$ (i)
   • $5x + 5y = 17.50$ (ii)
3. Multiply (ii) by $3/5$ to match the $y$ coefficient: $3x + 3y = 10.50$ (iii).
4. Subtract (iii) from (i): $(7x + 3y) - (3x + 3y) = 16.50 - 10.50$, resulting in $4x = 6$, so $x = 1.5$.
5. Substitute $x = 1.5$ into (iii): $3(1.5) + 3y = 10.50 \implies 4.5 + 3y = 10.50 \implies 3y = 6 \implies y = 2$.

The cost of 10 litres of red paint is $10 \times £2 = £20$.