Systems of Linear Equations

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Systems of Linear Equations · Grade 10 · Worksheet 3

1. Mere is designing a geometric pattern on a coordinate grid. She draws a triangle with vertices at A(10, 20), B(30, 50), and C(50, 10). She then draws the three medians of this triangle (lines from each vertex to the midpoint of the opposite side). Visualize this triangle and its medians on the grid. Write the system of linear equations representing two of these medians (the median from A to side BC and the median from B to side AC), then find the coordinates of their intersection point (the centroid). Answer: (30, 80/3) Solution: Find the midpoint of BC. B(30, 50) and C(50, 10). Midpoint M_BC = ((30+50)/2, (50+10)/2) = (80/2, 60/2) = (40, 30).
   Full step-by-step solution

   Step 1: Find the midpoint of BC. B(30, 50) and C(50, 10). Midpoint M_BC = ((30+50)/2, (50+10)/2) = (80/2, 60/2) = (40, 30). Step 2: Find the equation of the median from A(10, 20) to M_BC(40, 30). Slope m1 = (30 - 20) / (40 - 10) = 10/30 = 1/3. Using point-slope form with A(10, 20): y - 20 = (1/3)(x - 10). Multiply by 3: 3y - 60 = x - 10 => 3y = x + 50 => y = (1/3)x + 50/3. (Equation 1) Step 3: Find the midpoint of AC. A(10, 20) and C(50, 10). Midpoint M_AC = ((10+50)/2, (20+10)/2) = (60/2, 30/2) = (30, 15). Step 4: Find the equation of the median from B(30, 50) to M_AC(30, 15). Slope m2 = (15 - 50) / (30 - 30) = -35/0, which is undefined. This is a vertical line: x = 30. (Equation 2) Step 5: Solve the system: y = (1/3)x + 50/3 and x = 30. Substitute x = 30 into Equation 1: y = (1/3)(30) + 50/3 = 10 + 50/3 = 30/3 + 50/3 = 80/3. Step 6: The intersection point (centroid) is (30, 80/3).

2. On a coordinate plane, Emma draws a parallelogram with vertices at A(1, 1), B(7, 1), C(9, 5), and D(3, 5). She then draws a diagonal from A to C. A second line passes through the midpoint of this diagonal and is parallel to side AB. Visualize this figure. Write the system of linear equations representing the two lines (the diagonal AC and the line through the midpoint parallel to AB), then find their intersection point. Answer: (5, 3) Solution: Find the equation of diagonal AC. A(1, 1) and C(9, 5). Slope m1 = (5 - 1) / (9 - 1) = 4/8 = 1/2.
   Full step-by-step solution

   Step 1: Find the equation of diagonal AC. A(1, 1) and C(9, 5). Slope m1 = (5 - 1) / (9 - 1) = 4/8 = 1/2. Using point-slope form with A(1, 1): y - 1 = (1/2)(x - 1) => y = (1/2)x + 1/2. Multiply by 2: 2y = x + 1, or x - 2y = -1. (Equation 1) Step 2: Find midpoint M of AC. M = ((1+9)/2, (1+5)/2) = (10/2, 6/2) = (5, 3). Step 3: Find slope of AB. A(1, 1) and B(7, 1). Slope m2 = (1 - 1) / (7 - 1) = 0/6 = 0. So AB is horizontal (y = 1). The line through M parallel to AB is also horizontal, so its equation is y = 3. (Equation 2) Step 4: Solve the system: x - 2y = -1 and y = 3. Substitute y = 3 into x - 2(3) = -1 => x - 6 = -1 => x = 5. Step 5: The intersection point is (5, 3). This is exactly the midpoint M, so the two lines intersect at the midpoint of diagonal AC. The answer is (5, 3).

3. Solve: 7x + 3y = 31; 5x - 3y = 11; x = ? Answer: 3.5 Solution: Add the two equations to eliminate y: (7x + 3y) + (5x - 3y) = 31 + 11 → 12x = 42. Step 2: Divide both sides by 12: x = 42/12 = 7/2 = 3.5.
   Full step-by-step solution

   Step 1: Add the two equations to eliminate y: (7x + 3y) + (5x - 3y) = 31 + 11 → 12x = 42. Step 2: Divide both sides by 12: x = 42/12 = 7/2 = 3.5. The answer is x = 3.5.

4. Kaia is constructing a geometric design on a coordinate plane. She draws a triangle with vertices at A(9, 33), B(33, 9), and C(57, 33). She then draws the median from vertex B to side AC, and the median from vertex C to side AB. Visualize this triangle and its two medians. Write the system of linear equations representing these two medians, and find the coordinates of their intersection point. Answer: (33, 25) Solution: Find the midpoint of side AC. A(9, 33) and C(57, 33). Midpoint M_AC = ((9+57)/2, (33+33)/2) = (66/2, 66/2) = (33, 33).
   Full step-by-step solution

   Step 1: Find the midpoint of side AC. A(9, 33) and C(57, 33). Midpoint M_AC = ((9+57)/2, (33+33)/2) = (66/2, 66/2) = (33, 33). Step 2: Find the equation of the median from B(33, 9) to M_AC(33, 33). Slope m1 = (33 - 9) / (33 - 33) = 24/0, which is undefined. This is a vertical line: x = 33. (Equation 1) Step 3: Find the midpoint of side AB. A(9, 33) and B(33, 9). Midpoint M_AB = ((9+33)/2, (33+9)/2) = (42/2, 42/2) = (21, 21). Step 4: Find the equation of the median from C(57, 33) to M_AB(21, 21). Slope m2 = (21 - 33) / (21 - 57) = (-12) / (-36) = 1/3. Using point-slope form with C(57, 33): y - 33 = (1/3)(x - 57). Multiply by 3: 3y - 99 = x - 57. Simplify: 3y = x + 42, so y = (1/3)x + 14. (Equation 2) Step 5: Solve the system: x = 33 and y = (1/3)x + 14. Substitute x = 33 into Equation 2: y = (1/3)(33) + 14 = 11 + 14 = 25. Step 6: The intersection point is (33, 25). This is the centroid of the triangle.

5. Matiu is drawing a geometric design on a coordinate grid. He plots a parallelogram with vertices at A(2, 4), B(12, 4), C(16, 12), and D(6, 12). He then draws the diagonal from B to D. Next, he draws a line through the midpoint of diagonal BD that is parallel to side BC. Visualize this figure. Write the system of linear equations representing the diagonal BD and the line through its midpoint parallel to BC, then find their intersection point. Answer: (14, 8) Solution: Find the equation of diagonal BD. Points B(12, 4) and D(6, 12). Slope m1 = (12 - 4) / (6 - 12) = 8 / (-6) = -4/3.
   Full step-by-step solution

   Step 1: Find the equation of diagonal BD. Points B(12, 4) and D(6, 12). Slope m1 = (12 - 4) / (6 - 12) = 8 / (-6) = -4/3. Using point-slope form with B(12, 4): y - 4 = (-4/3)(x - 12). Multiply both sides by 3: 3y - 12 = -4(x - 12) => 3y - 12 = -4x + 48 => 4x + 3y = 60. (Equation 1) Step 2: Find midpoint M of BD. M = ((12 + 6)/2, (4 + 12)/2) = (18/2, 16/2) = (9, 8). Step 3: Find slope of BC. B(12, 4) and C(16, 12). Slope m2 = (12 - 4) / (16 - 12) = 8 / 4 = 2. The line through M parallel to BC has the same slope, 2. Using point-slope form: y - 8 = 2(x - 9) => y - 8 = 2x - 18 => y = 2x - 10. (Equation 2) Step 4: Solve the system: 4x + 3y = 60 and y = 2x - 10. Substitute y from Equation 2 into Equation 1: 4x + 3(2x - 10) = 60 => 4x + 6x - 30 = 60 => 10x = 90 => x = 9. Step 5: Substitute x = 9 into y = 2x - 10: y = 2(9) - 10 = 18 - 10 = 8. Step 6: The intersection point is (9, 8). This is exactly the midpoint M of diagonal BD. The answer is (9, 8).

6. Aroha and Tane are volunteering at a local animal shelter. They are preparing two types of care packages for adopted pets: a standard care package and a premium care package. Each standard package contains 3 bags of food and 1 toy, while each premium package contains 5 bags of food and 3 toys. The shelter has exactly 63 bags of food and 33 toys available. Kaia, the shelter coordinator, wants to use all the food and all the toys exactly, with nothing left over. How many of each type of care package should they prepare? Answer: standard: 12, premium: 7 Solution: Let s = number of standard packages, p = number of premium packages. Food equation: 3s + 5p = 63 Toy equation: 1s + 3p = 33 Solve using elimination.
   Full step-by-step solution

   Step 1: Let s = number of standard packages, p = number of premium packages. Step 2: Food equation: 3s + 5p = 63 Step 3: Toy equation: 1s + 3p = 33 Step 4: Solve using elimination. Multiply the toy equation by 3 to align s coefficients: 3s + 9p = 99 Step 5: Subtract the food equation from this result: (3s + 9p) - (3s + 5p) = 99 - 63 Step 6: Simplify: 4p = 36, so p = 9 Step 7: Substitute p = 9 into the toy equation: s + 3(9) = 33, so s + 27 = 33, thus s = 6 Step 8: Check with food equation: 3(6) + 5(9) = 18 + 45 = 63. Correct. The answer is 6 standard packages and 9 premium packages.