Solving Problems Involving Linear Inequalities in Two Variables PDF

# Solving Problems Involving Linear Inequalities in Two Variables ## Key Words Use an inequality appropriate for the condition of word problems. | Inequality | Phrase | |---|---| | `<3` | Less than | | `>3` | More than, Greater than | | `≤3` | At most, No more than, Not more than, Not exceed | |...

# Solving Problems Involving Linear Inequalities in Two Variables ## Key Words Use an inequality appropriate for the condition of word problems. | Inequality | Phrase | |---|---| | `<3` | Less than | | `>3` | More than, Greater than | | `≤3` | At most, No more than, Not more than, Not exceed | | `≥3` | At least | ## Steps In Solving Problems 1. Understand the problem 2. Plan and represent the given 3. Carry out the plan 4. Check the result ## Example 1. Solve the following problems involving linear inequalities in two variables. 1. What are the values of $y$ in $y < 6 - 3x$ if $x = -5$? $y < 6 - 3(-5)$ $y < 6 + 15$ $y < 21$ $(-∞, 21)$ **Set Notation** 2. What are the values of $x$ in $3x + 1 ≤ 7y$ if $y = -2$? $3x + 1 ≤ 7(-2)$ $3x + 1 ≤ -14$ $3x ≤ -14 - 1 $ $3x≤ -15$ $x ≤ -5$ $(-∞,-5]$ **Set Notation** 3. Find the range of values of $y$ in the equation $y = 2x - 3$ for $-1 < x ≤ 1$. $y = 2x - 3$ $y = 2(-1) - 3$ $y = -2 - 3$ $y = -5$ $y = 2x - 3$ $y = 2(1)-3$ $y = 2 - 3$ $y = -1$ $-5 < y ≤ -1$ **Take note that the inequality symbols should match the ones in the given range, and the order must be correct.** ## Example 2. Alex, a working student, has two part-time jobs, one paying Php 30 per hour and another is paying Php 28 per hour. He must earn at least Php 840 per week to pay his expenses while attending college night classes. Write an inequality that shows the various ways he can schedules his time to achieve his goal. | Hourly rate on the first job | Hours worked on the first job | Hourly rate on the second job | Hours worked on the second job| | |---|---|---|---|---| | 30 | *x* | 28 | *y* | ≥ 840 | The inequality obtained is $30x + 28y ≥ 840$. ## Example 2 (cont.) Write an inequality that shows the various ways he can schedules his time to achieve his goal. The inequality obtained is $30x + 28y ≥ 840$. * $(10,15)$ - $30(10) + 28(15) ≥ 840$ - $300 + 420 ≥ 840$ - $720 ≥ 840$ - **FALSE** * $(40,30)$ - $30(40) + 28(30) ≥ 840$ - $1200 + 840 ≥ 840$ - $2040 > 840$ - **TRUE** * $(28,0)$ - $30(28) + 28(0) ≥ 840$ - $840 + 0 ≥ 840$ - $840 ≥ 840$ - **TRUE** ## Example 3. A clinic is writing a budget plan for the yearly inventory so they can buy additional supplies. They estimate that the cost a set of face mask is P 80 per box, and P 120 per half-gallon bottle of alcohol. They also plan to spend not more than P 2400. (Let x = boxes of face masks, y = bottles of alcohol) 1. Write an inequality that best describes the problem? | Price of set of face mask | Number of boxes of face masks | Price of half-gallon bottle of alcohol | Number of half-gallon bottle of alcohol| | |---|---|---|---|---| | 80 | *x* | 120 | *y* | ≤ 2400 | The inequality obtained is $80x + 120y ≤ 2400$. ## Example 3 (cont.) 2. Up to how many bottles of alcohol can be bought if they did not order for face masks? $80x + 120y ≤ 2400$ *x* = boxes of face masks, *y* = bottles of alcohol $(0, y)$ $80(0) + 120y ≤ 2400$ $120y ≤ 2400$ $y ≤ 20$ 3. If the clinic opted to buy 4 boxes of face masks, how many bottles of alcohol can be purchased? $(4, y)$ $80(4) + 120y ≤ 2400$ $320 + 120y ≤ 2400$ $120y ≤ 2400 - 320$ $120y ≤ 2080$ $y ≤ 17.33$ $y ≤ 17$ 4. How much can they save if they bought 10 items each? *x* = boxes of face masks, *y* = bottles of alcohol $(10,10)$ $80(10) + 120(10) ≤ 2400$ $800 + 1200 ≤ 2400$ $2000 ≤ 2400$ Php 2400 - Php 2000 = Php 400 They can save Php 400 if they bought 10 items each. ## Example 4. Mr. Ben bought 4kg of soya beans and 5kg of brown sugar, he paid not more than Php 524. If a kilogram of brown sugar is Php 40, then how much is the cost of soya beans? 1. Represents the unknown by using variables Let *x* = Soya beans Let *y* = brown sugar 2. Write a linear inequality $4x + 5y ≤ 524$ 3. Solve the linear inequality How much is the cost of soya beans? Since a kilogram of brown sugar cost Php 40, let *y* = 40 $4x + 5(40) ≤ 524$ $4x + 200 ≤ 524$ $4x ≤ 524 - 200$ $4x ≤ 324$ $x ≤ \frac{324}{4}$ $x ≤ 81$ Therefore, soya beans cost less than or equal to Php 81. ## Example 5. Ana needs to buy fruits. An orange cost Php 15 each while an apple cost Php 10 each. She spends no more than Php 150. The total number of fruits should not exceed 10. 1. Represents the unknown by using variables Let *x* = the number of orange Let *y* = the number of apple 2. Write a system of linear inequalities * An orange cost Php 15 each while an apple cost Php 10 each. She spends no more than Php 150 | Cost of Orange | Times | Number of orange | Plus | Cost of Apple | Times | Number of Apple | No More than | Php 150| |---|---|---|---|---|---|---|---|---| | 15 | *.* | *x* | + | 10 | *.* | *y* | ≤ | 150| $15x + 10y ≤ 150$ * The total number of fruits should not exceed 10 | Number of orange | Plus | Number of Apple | Not exceed | Php 10| |---|---|---|---|---| | *x* | + | *y* | ≤ | 10 | $x + y ≤ 10$ 3. Solve system of linear inequalities by graphing. { $15x + 10y ≤ 150$ $x + y ≤ 10$ } Replace the inequality symbol with an equal sign: $15x + 10y ≤ 150$ - $15x + 10y = 150$ $x + y ≤ 10$ - $x + y = 10$ Find the x-intercept and y-intercept: $15x + 10y = 150$ * Let *x* = 0 - $15(0) + 10y = 150$ - $10y = 150$ - $y = \frac{150}{10}$ - $y = 15$ * Let *y* = 0 - $15x + 10(0) = 150$ - $15x = 150$ - $x = \frac{150}{15}$ - $x = 10$ $x + y = 10$ * Let *x* = 0 - $(0) + y = 10$ - $y=10$ * Let *y* = 0 - $x + (0) = 10$ - $x = 10$ **Graph**: | Points | 15x + 10y ≤ 150 | x + y ≤ 10 | |---|---|---| | (0, 15) | | | | (10, 0) | | | | (0, 10) | | | | (10, 0) | | | **Use the origin as a testing point (0,0)** $15x + 10y ≤ 150$ - $15(0) + 10(0) ≤ 150$ - $0 + 0 ≤ 150$ - $0 ≤ 150$ - **True** **Shade the area where the origin is located** $x + y ≤ 10$ - $(0) + (0) ≤ 10$ - $0 + 0 ≤ 10$ - $0 ≤ 10$ - **True** **Shade the area where the origin is located** **Therefore, the solution set is the double shaded region.** There are many combinations but 3 of them are: * 3 oranges * 1 apple * 2 oranges * 5 apples * 1 orange * 2 apples 4. Check the solution set Let's obtain 3 points from the double shaded region: | Points | 15x + 10y ≤ 150 | x + y = 10 | |---|---|---| | (3,1) | $15(3) + 10(1) ≤ 150$ - $45 + 10 ≤ 150$ - $55 ≤ 150$ - **True** | $(3)+(1) ≤ 10$ - $3 + 1 ≤ 10$ - $4 ≤ 10$ - **True** | | (1, 2) | $15(1) + 10(2) ≤ 150$ - $15 + 20 ≤ 150$ - $35 ≤ 150$ - **True** | $(1)+(2) ≤ 10$ - $1 + 2 ≤ 10$ - $3 ≤ 10$ - **True** | | (2, 5) | $15(2) + 10(5) ≤ 150$ - $30 + 50 ≤ 150$ - $80 ≤ 150$ - **True** | $(2)+(5) ≤ 10$ - $2 + 5 ≤ 10$ - $7 ≤ 10$ - **True** |