Equations and Inequalities

Lessons

Senior 4 · Basic Mathematics

Equations and Inequalities

5 lessons

Lesson 4 of 5

Simultaneous linear equations and inequalities

25 min

Two prices from two receipts

At a market stall, Mutesi pays 5 700 Frw for 2 kg of beans and 3 kg of rice. Her neighbour pays 4 900 Frw for 4 kg of beans and 1 kg of rice from the same stall (illustrative prices). Neither receipt alone tells you the price of a kilogram of beans, $b$ , or of rice, $r$ : each gives one equation with two unknowns, and many pairs $(b, r)$ fit it. But only one pair fits both:

\begin{cases} 2b + 3r = 5700 \\ 4b + r = 4900 \end{cases}

Two equations that must hold at the same time form a system of simultaneous equations. This lesson solves them in three ways, and then moves to systems of inequalities, which describe all the plans a business can afford.

Definition

System of two linear equations in two unknowns: $\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$ . A solution is an ordered pair $(x, y)$ that satisfies both equations.

Each equation is a straight line, so a solution is a point where the two lines meet. Two lines can cross once (one solution), be parallel (no solution) or be the same line (infinitely many solutions).

Example: $(2, 3)$ solves $x + y = 5$ and $2 x - y = 1$ , since $2 + 3 = 5$ and $4 - 3 = 1$ .

Worked example 1: substitution

Solve the market system $2 b + 3 r = 5700$ , $4 b + r = 4900$ .

Solution

The second equation gives $r$ easily: $r = 4900 - 4 b$ . Substitute into the first:

2b + 3(4900 - 4b) = 5700 \;\Longrightarrow\; 2b + 14\,700 - 12b = 5700 \;\Longrightarrow\; -10b = -9000 \;\Longrightarrow\; b = 900 .

Then $r = 4900 - 4 (900) = 1300$ . Beans cost 900 Frw per kg and rice 1 300 Frw per kg. Check: $2 (900) + 3 (1300) = 1800 + 3900 = 5700$ . ✓

Worked example 2: equating one variable (comparison)

Solve $x + y = 5$ and $2 x - y = 1$ by making $y$ the subject of both.

Solution

$y = 5 - x$ and $y = 2 x - 1$ . Both equal $y$ , so they equal each other: $5 - x = 2 x - 1$ , $6 = 3 x$ , $x = 2$ , and $y = 5 - 2 = 3$ . The solution is $(2, 3)$ .

Worked example 3: elimination

Solve $3 x + 2 y = 16$ and $5 x - 3 y = 14$ .

Solution

Make the $y$ -coefficients opposite: multiply the first equation by 3 and the second by 2.

\begin{aligned} 9x + 6y &= 48 \\ 10x - 6y &= 28 \end{aligned}

Add: $19 x = 76$ , so $x = 4$ . Then $3 (4) + 2 y = 16$ gives $y = 2$ . Check in the second equation: $20 - 6 = 14$ . ✓ $S = \{(4, 2)\}$ .

Solving graphically

Draw both lines on one grid (two points for each line are enough) and read the point where they cross. For $x + y = 5$ use $(0, 5)$ and $(5, 0)$ ; for $2 x - y = 1$ use $(0, - 1)$ and $(3, 5)$ .

A graph shows the meaning of the solution at a glance, but it is only as accurate as the drawing, so confirm it algebraically. The graph also explains the special cases: $x + y = 5$ and $x + y = 7$ are parallel lines (no solution); $x + y = 5$ and $2 x + 2 y = 10$ are the same line (infinitely many solutions).

Warning

Multiply the whole equation. When you multiply an equation to match coefficients, multiply every term, including the right-hand side: $3 \times (3x + 2y = 16)$ is $9 x + 6 y = 48$ , not $9 x + 6 y = 16$ . And always check your pair in the equation you did not use last.

Simultaneous linear inequalities in two unknowns

A line $a x + b y = c$ cuts the plane into two half-planes: on one side $ax + by \lt c$ , on the other $ax + by \gt c$ . To draw $ax + by \le c$ : draw the boundary line (solid for $\leq \le$ or $\geq \ge$ , dashed for $< \lt$ or $> \gt$ ), then test one point not on the line, usually $(0, 0)$ . If the point satisfies the inequality, its side is the solution; if not, the other side is. The solution of several inequalities is the region where all their half-planes overlap.

Worked example 4: a tailor's possible weekly plans

A tailor makes $x$ shirts and $y$ dresses a week. A shirt needs 2 m of cloth and 2 hours; a dress needs 3 m of cloth and 1 hour. She has at most 24 m of cloth and 16 hours a week (illustrative). Write the conditions, draw the region of possible plans, and decide whether 5 shirts and 3 dresses, or 7 shirts and 3 dresses, are possible.

Solution

\begin{cases} 2x + 3y \le 24 & \text{(cloth)} \\ 2x + y \le 16 & \text{(time)} \\ x \ge 0,\ y \ge 0 \end{cases}

Testing $(0, 0)$ : $0 \le 24$ and $0 \le 16$ are true, so both regions lie on the origin's side of their lines. The two boundary lines meet where $2 x + 3 y = 24$ and $2 x + y = 16$ : subtracting gives $2 y = 8$ , so $y = 4$ and $x = 6$ .

$(5, 3)$ : $10 + 9 = 19 \le 24$ and $10 + 3 = 13 \le 16$ : possible. $(7, 3)$ : $14 + 9 = 23 \le 24$ but $14 + 3 = 17 \gt 16$ : not enough time, so not possible.

Tip

Choose the method by the coefficients. If one unknown already has coefficient 1 or $- 1$ , substitute. If not, eliminate. Use a graph when the question asks for one, or when you need to see a region rather than a single point.

Brian McLogan: Solve a System by Graphing, Substitution, Elimination. Watch the same system solved three ways, giving the same point each time.

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Practice

Try each one before opening the answers. They get harder as you go.

Solve by substitution: $y = 3 x - 1$ , $2 x + y = 9$ .
Solve by elimination: $4 x + 3 y = 10$ , $2 x - 3 y = 14$ .
Solve by elimination: $5 x + 2 y = 1$ , $3 x + 4 y = - 5$ .
Solve graphically, then check algebraically: $y = x + 1$ , $x + y = 7$ .
Show that $2 x - y = 3$ and $6 x - 3 y = 5$ have no solution, and explain what this means for the lines.
Tickets for a school concert cost 1 500 Frw for students and 3 000 Frw for adults. 220 tickets were sold for 465 000 Frw. How many adults came?
Draw the region $x \ge 0$ , $y \ge 0$ , $x + 2y \le 10$ , $3x + y \le 15$ . Find its corners. Is $(3, 3)$ in it?

Show answersHide answers

$2 x + 3 x - 1 = 9$ , $x = 2$ , $y = 5$ .
Add: $6 x = 24$ , $x = 4$ ; $16 + 3 y = 10$ , $y = - 2$ .
Double the first: $10 x + 4 y = 2$ ; subtract the second: $7 x = 7$ , $x = 1$ , $y = - 2$ .
$x + 1 = 7 - x$ , $x = 3$ , $y = 4$ : the lines cross at $(3, 4)$ .
Tripling the first gives $6 x - 3 y = 9$ , which contradicts $6 x - 3 y = 5$ . The lines are parallel (same gradient 2), so they never meet.
$s + a = 220$ , $1500s + 3000a = 465\,000$ . Then $1500(220 - a) + 3000a = 465\,000$ , $1500a = 135\,000$ , $a = 90$ adults (and 130 students).
Corners $(0, 0)$ , $(5, 0)$ , $(4, 3)$ , $(0, 5)$ . $(3, 3)$ : $3 + 6 = 9 \le 10$ and $9 + 3 = 12 \le 15$ , so yes.

Quick check

Which pair solves x + 2y = 8 and x − y = 2?

Key points

A solution of a system satisfies every equation: it is where the lines cross.
Substitution or comparison when a coefficient is 1; elimination otherwise; graphs to see the meaning.
Parallel lines: no solution; the same line: infinitely many.
For inequalities in $x$ and $y$ : draw each boundary, test $(0, 0)$ , and take the region where all conditions hold.

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the questions. All prices and the tailor's figures are illustrative.
Video: Solve a System by Graphing, Substitution, Elimination, Brian McLogan, embedded from YouTube.

1Two prices from two receipts

2Solving graphically

3Simultaneous linear inequalities in two unknowns

Two prices from two receipts

Solving graphically

Simultaneous linear inequalities in two unknowns