Simultaneous linear inequalities

Unit 5 Simultaneous linear inequalities

Key unit competence: Solve real-world problems using systems of linear equations and inequalities.

Key ideas

• A compound inequality like $-2<x\le 5$ means $x>-2$ and $x\le 5$. Solve each part, then keep the values satisfying both (the intersection).
• "or" statements keep values satisfying at least one part (union).
• Sign table of a binomial $ax+b$: it is zero at $x=-\frac{b}{a}$, has the sign of $a$ to the right of this root and the opposite sign to the left.
• To solve $(ax+b)(cx+d)>0$ or $\frac{ax+b}{cx+d}\ge 0$: find the roots, make a sign table, choose the intervals with the right sign. For a quotient, the root of the denominator is always excluded.
• The graph of $ax+by=c$ is a line. The inequality $ax+by<c$ is a half-plane on one side of it: test a point such as $(0,0)$. Draw a dashed line for $<$, $>$ and a solid line for $\le$, $\ge$.
• A system of inequalities in two unknowns is solved by graphing each half-plane; the solution is the region common to all (the feasible region). Shade the unwanted regions to leave the solution region clear.

Formulas to know

Common misconceptions

• "$(x-1)(x-4)>0$ gives $x>1$ and $x>4$." No: use a sign table: $x<1$ or $x>4$.
• "$\frac{x-1}{x+2}\ge 0$ includes $x=-2$." No: $x=-2$ makes the denominator zero; it is always excluded.
• "The line $y=2x+1$ is part of the region $y<2x+1$." No: for a strict inequality the line is dashed and not included.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Solve and show on a number line: $-3<2x-1\le 5$.

   Show answer

   $-2<2x\le 6$, so $-1<x\le 3$: open circle at $-1$, closed circle at 3, shading between.

   Common mistake: Using closed circles at both ends.

2. Find all integers $x$ such that $x^2-5x+4<0$.

   Show answer

   $(x-1)(x-4)<0$: $1<x<4$, integers 2 and 3.

   Common mistake: Including 1 and 4.

3. Make a table of signs for $(x-2)(x+1)$ and solve $(x-2)(x+1)\ge 0$.

   Show answer

   Roots $-1$, 2. Positive for $x<-1$, negative between, positive for $x>2$. Solution: $x\le -1$ or $x\ge 2$.

   Common mistake: Taking the region between the roots.

4. Solve the system $x-y<1$, $x+y>3$ and give one point of the solution region.

   Show answer

   Region above $y=x-1$ (dashed) and above $y=-x+3$ (dashed). Example: $(2,2)$: $0<1$ and $4>3$.

   Common mistake: Drawing solid lines for strict inequalities.

5. Solve $\frac{x-3}{x+1}\le 0$.

   Show answer

   $-1<x\le 3$ ($x=-1$ excluded).

   Common mistake: Including $x=-1$.

6. A farmer has 12 ha for beans ($x$) and maize ($y$). He wants at least 3 ha of beans and at least twice as much maize as beans. Write the inequalities.

   Show answer

   $x+y\le 12$, $x\ge 3$, $y\ge 2x$, $y\ge 0$.

   Common mistake: Writing $x\ge 2y$ for "twice as much maize as beans".

7. Show on a graph the region $y\le x+2$, $y\ge 0$, $x\le 3$ and list its corner points.

   Show answer

   Corners $(-2,0)$, $(3,0)$, $(3,5)$ (the region is a triangle under $y=x+2$, above the x-axis, left of $x=3$).

   Common mistake: Testing the wrong side of each line.

8. Write the inequality for the half-plane below the line through $(0,2)$ and $(4,0)$, including the line.

   Show answer

   The line is $x+2y=4$; below it: $x+2y\le 4$.

   Common mistake: Getting the line's equation wrong ($y=2x+4$).

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Solve $2x-1>x+3$ and $3x-2<2x+5$.	a. the side containing the origin is the solution
   2. To check which side of $2x+y=4$ satisfies $2x+y<4$, test $(0,0)$. The result is:	b. $4<x<7$
   3. Which point satisfies $x+y<5$ and $x>1$?	c. $(2,1)$
   4. Solve $3x-2<x+4<2x+6$.	d. $-2<x<3$

   Show answer

   1 → b ($4<x<7$) · 2 → a (the side containing the origin is the solution) · 3 → c ($(2,1)$) · 4 → d ($-2<x<3$)

   • Step 1. First: $x>4$. Second: $x<7$.
     Step 2. Both must hold: $4<x<7$.
   • $0<4$ is true.
   • $2+1<5$ and $2>1$.
   • $2x<6$ gives $x<3$; $-2<x$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   15 shirts and 10 trousers  ·  6  ·  4  ·  the region satisfying all inequalities
   1. The feasible region of a system of inequalities is .
   2. A shop stocks $x$ shirts and $y$ trousers with $x\ge 10$, $y\ge 5$ and $x+y\le 30$. Which is possible? →
   3. How many integer pairs $(x,y)$ satisfy $x\ge 1$, $y\ge 1$, $x+y\le 4$? →
   Show answer

   1: the region satisfying all inequalities · 2: 15 shirts and 10 trousers · 3: 6

   • It is the intersection of the half-planes.
   • $15\ge 10$, $10\ge 5$, $25\le 30$.
   • $(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Solve $x<2$ or $x>5$. Which number is a solution?	a. $2<x<6$
   2. Solve $-1\le 2x+1<7$.	b. $-1\le x<3$
   3. Solve $(2x-1)(x+3)\le 0$.	c. 7
   4. Solve $x+3>5$ and $2x<12$.	d. $-3\le x\le \frac{1}{2}$

   Show answer

   1 → c (7) · 2 → b ($-1\le x<3$) · 3 → d ($-3\le x\le \frac{1}{2}$) · 4 → a ($2<x<6$)

   • 7 satisfies $x>5$.
   • Subtract 1: $-2\le 2x<6$; divide by 2.
   • Step 1. Critical values: $x=-3$ and $x=\frac{1}{2}$.
     Step 2. The product is negative between them and zero at them: $-3\le x\le \frac{1}{2}$.
   • $x>2$ and $x<6$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $x<1$ or $x>4$  ·  13  ·  5  ·  6
   1. Solve $(x-1)(x-4)>0$. →
   2. Find the largest integer $x$ with $3(x-2)<x+5$. →
   3. Find the greatest value of $3x+2y$ for integers with $x\ge 0$, $y\ge 0$, $x+y\le 5$, $x\le 3$. →
   Show answer

   1: $x<1$ or $x>4$ · 2: 5 · 3: 13

   • The product is positive outside the roots.
   • Step 1. $3x-6<x+5\Rightarrow 2x<11\Rightarrow x<5.5$.
     Step 2. The largest integer below 5.5 is 5.
   • Best at $(3,2)$: $9+4=13$ (corner points: $(0,5)$ gives 10, $(3,0)$ gives 9).

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