Year 10AC9M10A02

Simultaneous Equations from a Graph

Solve linear inequalities and simultaneous linear equations in 2 variables; interpret solutions graphically and communicate solutions in terms of the situation.

Two lines, one crossing point. That point is the only pair of values that satisfies both equations at once — which is exactly what solving simultaneously means.

Builds onYr 9 · Simultaneous Equations

Method

The solution is the intersection

Where the lines cross, both equations are true. That single point is the answer.

Parallel lines mean no solution

m_1 = m_2,\ c_1 \neq c_2

Same gradient, different intercept — they never meet, so no pair of values works.

The same line twice means infinitely many

m_1 = m_2,\ c_1 = c_2

Every point on the line satisfies both.

Always check by substituting back

Put your point into both original equations. Reading a graph is approximate; substitution is not.

Worked example

e.g.

\text{Solve graphically: } y = 2x - 1 \quad \text{and} \quad y = -x + 5

Write both equations in the form $y = mx + c$ if they are not already. $y = 2x - 1 \qquad y = -x + 5$
Draw both lines on the same axes using gradient and intercept.
Read off the point where they cross. $(2;3)$
Check it in the first equation. $2(2) - 1 = 3 \quad\checkmark$
Check it in the second. Both must work, or the point is wrong. $-(2) + 5 = 3 \quad\checkmark$
State the solution as a pair of values, not just a point. $x = 2,\quad y = 3$

Practice

Find the intersection, then substitute back into both equations.

y = x + 2 \ \text{ and } \ y = -x + 6

Answer

x = 2,\ y = 4

y = 3x \ \text{ and } \ y = x + 4

Answer

x = 2,\ y = 6

\text{How many solutions has } y = 2x + 1 \text{ and } y = 2x - 3?

Answer

\text{None — same gradient, different intercept, so the lines are parallel.}

\text{How many solutions has } y = 4x - 2 \text{ and } 2y = 8x - 4?

Answer

\text{Infinitely many — the second is the first doubled, so it is the same line.}

\text{You read the crossing point as } (3;5) \text{ but } 3 + 2 \neq 5 \text{ for } y = x + 2. \text{ What now?}

Answer

\text{The reading is wrong. Solve algebraically — a graph gives an estimate, substitution gives proof.}

\text{Two hire firms charge } y = 40 + 2x \text{ and } y = 25 + 5x. \text{ After how many units is the first cheaper?}

Answer

40 + 2x = 25 + 5x \ \Rightarrow\ x = 5 \quad\therefore\quad \text{cheaper beyond 5 units}