Please solve the following problems: 1. If 3x+5=20, find the value of x. 2. Solve: 2(x-3)+5=17. 3. Factorise: x2-5x+6. 4. Expand and simplify: (x+2)(x-3).

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Asked on May 19, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-05-19T21:03:42.352Z

You're on a roll — here are the solutions for all the problems from the image. Problem 1: If 3x+5=20, find the value of x. Step 1: Subtract 5 from both sides of the equation. 3x+5-5 = 20-5 3x = 15 Step 2: Divide both sides by 3. (3x)/(3) = (15)/(3) x = 5 The value of x is: 5 Problem 2: Solve: 2(x-3)+5=17. Step 1: Distribute the 2 into the parenthesis. 2x - 6 + 5 = 17 Step 2: Combine the constant terms on the left side. 2x - 1 = 17 Step 3: Add 1 to both sides of the equation. 2x - 1 + 1 = 17 + 1 2x = 18 Step 4: Divide both sides by 2. (2x)/(2) = (18)/(2) x = 9 The value of x is: 9 Problem 3: Factorise: x^2-5x+6. Step 1: Find two numbers that multiply to 6 and add up to -5. The numbers are -2 and -3, since (-2) × (-3) = 6 and (-2) + (-3) = -5. Step 2: Write the quadratic expression in factored form using these numbers. (x-2)(x-3) The factored form is: (x-2)(x-3) Problem 4: Expand and simplify: (x+2)(x-3). Step 1: Use the FOIL method (First, Outer, Inner, Last) to expand the product. (x+2)(x-3) = x · x + x · (-3) + 2 · x + 2 · (-3) = x^2 - 3x + 2x - 6 Step 2: Combine the like terms. = x^2 - x - 6 The expanded and simplified expression is: x^2-x-6 Problem 5: Solve the equation: x^2-4x-12=0. Step 1: Factor the quadratic expression. Find two numbers that multiply to -12 and add up to -4. The numbers are -6 and 2, since (-6) × 2 = -12 and (-6) + 2 = -4. (x-6)(x+2) = 0 Step 2: Set each factor equal to zero and solve for x. x-6 = 0 or x+2 = 0 x = 6 or x = -2 The solutions for x are: x=6 or x=-2 Problem 6: If = (3)/(5), find the value of . Step 1: Use the Pythagorean identity ^2 + ^2 = 1. Substitute the given value of : ((3)/(5))^2 + ^2 = 1 (9)/(25) + ^2 = 1 Step 2: Subtract (9)/(25) from both sides. ^2 = 1 - (9)/(25) ^2 = (25)/(25) - (9)/(25) ^2 = (16)/(25) Step 3: Take the square root of both sides. = ±sqrt((16)/(25)) = ±(4)/(5) Assuming is an acute angle (in the first quadrant), is positive. The value of is: (4)/(5) Problem 7: In ABC, A=50^, B=60^. Find C. Step 1: The sum of the angles in a triangle is 180^. A + B + C = 180^ Step 2: Substitute the given values for A and B. 50^ + 60^ + C = 180^ 110^ + C = 180^ Step 3: Subtract 110^ from both sides to find C. C = 180^ - 110^ C = 70^ The measure of C is: 70^ Problem 8: Find the area of a triangle with base 10 cm and height 6 cm. Step 1: Use the formula for the area of a triangle: Area = (1)/(2) × base × height. Area = (1)/(2) × 10 cm × 6 cm Step 2: Calculate the area. Area = (1)/(2) × 60 cm^2 Area = 30 cm^2 The area of the triangle is: 30 cm^2 Problem 9: The perimeter of a rectangle is 40 cm and its length is 14 cm. Find its breadth. Step 1: Use the formula for the perimeter of a rectangle: P = 2(l+b), where P is the perimeter, l is the length, and b is the breadth. Substitute the given values: P=40 cm and l=14 cm. 40 = 2(14+b) Step 2: Divide both sides by 2. (40)/(2) = 14+b 20 = 14+b Step 3: Subtract 14 from both sides to find b. 20 - 14 = b b = 6 cm The breadth of the rectangle is: 6 cm Problem 10: A bag contains 3 red balls, 4 blue balls and 5 green balls. One ball is drawn at random. What is the probability of getting a blue ball? Step 1: Calculate the total number of balls in the bag. Total balls = 3 (red) + 4 (blue) + 5 (green) = 12 balls Step 2: Identify the number of blue balls. Number of blue balls = 4 Step 3: Calculate the probability of drawing a blue ball using the formula: P(blue ball) = Number of blue ballsTotal number of balls P(blue ball) = (4)/(12) Step 4: Simplify the fraction. P(blue ball) = (1)/(3) The probability of getting a blue ball is: (1)/(3) Problem 11: Simplify: ((2x)/(3))^2. Step 1: Apply the square to both the numerator and the denominator. ((2x)/(3))^2 = ((2x)^2)/(3^2) Step 2: Calculate the squares. = (2^2 x^2)/(3^2) = (4x^2)/(9) The simplified expression is: (4x^2)/(9) Problem 12: Find the value of y if (y)/(2)+3=11. Step 1: Subtract 3 from both sides of the equation. (y)/(2)+3-3 = 11-3 (y)/(2) = 8 Step 2: Multiply both sides by 2. (y)/(2) × 2 = 8 × 2 y = 16 The value of y is: 16 Problem 13: Find the coordinates of the midpoint of the line segment joining (2,3) and (6,7). Step 1: Use the midpoint formula: M = ((x_1+x_2)/(2), (y_1+y_2)/(2)). Let (x_1, y_1) = (2,3) and (x_2, y_2) = (6,7). Step 2: Substitute the coordinates into the formula. M = ((2+6)/(2), (3+7)/(2)) Step 3: Calculate the sums and divide by 2. M = ((8)/(2), (10)/(2)) M = (4,5) The coordinates of the midpoint are: (4,5) Problem 14: In the figure, lines l and m are parallel. Find the value of x. Step 1: Identify the relationship between the angles. The angle 120^ and the angle adjacent to it on line l form a linear pair, so their sum is 180^. Let this adjacent angle be . + 120^ = 180^ = 180^ - 120^ = 60^ Step 2: Recognize that and x^ are corresponding angles because lines l and m are parallel. Corresponding angles are equal. x^ = x^ = 60^ The value of x is: 60 Problem 15: Prove that: (a+b)^2 = a^2+2ab+b^2. Step 1: Start with the left-hand side of the equation. (a+b)^2 Step 2: Expand the square as a product of two binomials. (a+b)^2 = (a+b)(a+b) Step 3: Use the distributive property (FOIL method) to multiply the binomials. (a+b)(a+b) = a · a + a · b + b · a + b · b = a^2 + ab + ba + b^2 Step 4: Combine the like terms (ab and ba are the same). = a^2 + 2ab + b^2 This matches the right-hand side of the equation, thus the identity is proven. (a+b)^2 = a^2+2ab+b^2 All done

Accepted Answer · 2026-05-19T21:03:42.352Z

You're on a roll — here are the solutions for all the problems from the image. Problem 1: If 3x+5=20, find the value of x. Step 1: Subtract 5 from both sides of the equation. 3x+5-5 = 20-5 3x = 15 Step 2: Divide both sides by 3. (3x)/(3) = (15)/(3) x = 5 The value of x is: 5 Problem 2: Solve: 2(x-3)+5=17. Step 1: Distribute the 2 into the parenthesis. 2x - 6 + 5 = 17 Step 2: Combine the constant terms on the left side. 2x - 1 = 17 Step 3: Add 1 to both sides of the equation. 2x - 1 + 1 = 17 + 1 2x = 18 Step 4: Divide both sides by 2. (2x)/(2) = (18)/(2) x = 9 The value of x is: 9 Problem 3: Factorise: x^2-5x+6. Step 1: Find two numbers that multiply to 6 and add up to -5. The numbers are -2 and -3, since (-2) × (-3) = 6 and (-2) + (-3) = -5. Step 2: Write the quadratic expression in factored form using these numbers. (x-2)(x-3) The factored form is: (x-2)(x-3) Problem 4: Expand and simplify: (x+2)(x-3). Step 1: Use the FOIL method (First, Outer, Inner, Last) to expand the product. (x+2)(x-3) = x · x + x · (-3) + 2 · x + 2 · (-3) = x^2 - 3x + 2x - 6 Step 2: Combine the like terms. = x^2 - x - 6 The expanded and simplified expression is: x^2-x-6 Problem 5: Solve the equation: x^2-4x-12=0. Step 1: Factor the quadratic expression. Find two numbers that multiply to -12 and add up to -4. The numbers are -6 and 2, since (-6) × 2 = -12 and (-6) + 2 = -4. (x-6)(x+2) = 0 Step 2: Set each factor equal to zero and solve for x. x-6 = 0 or x+2 = 0 x = 6 or x = -2 The solutions for x are: x=6 or x=-2 Problem 6: If = (3)/(5), find the value of . Step 1: Use the Pythagorean identity ^2 + ^2 = 1. Substitute the given value of : ((3)/(5))^2 + ^2 = 1 (9)/(25) + ^2 = 1 Step 2: Subtract (9)/(25) from both sides. ^2 = 1 - (9)/(25) ^2 = (25)/(25) - (9)/(25) ^2 = (16)/(25) Step 3: Take the square root of both sides. = ±sqrt((16)/(25)) = ±(4)/(5) Assuming is an acute angle (in the first quadrant), is positive. The value of is: (4)/(5) Problem 7: In ABC, A=50^, B=60^. Find C. Step 1: The sum of the angles in a triangle is 180^. A + B + C = 180^ Step 2: Substitute the given values for A and B. 50^ + 60^ + C = 180^ 110^ + C = 180^ Step 3: Subtract 110^ from both sides to find C. C = 180^ - 110^ C = 70^ The measure of C is: 70^ Problem 8: Find the area of a triangle with base 10 cm and height 6 cm. Step 1: Use the formula for the area of a triangle: Area = (1)/(2) × base × height. Area = (1)/(2) × 10 cm × 6 cm Step 2: Calculate the area. Area = (1)/(2) × 60 cm^2 Area = 30 cm^2 The area of the triangle is: 30 cm^2 Problem 9: The perimeter of a rectangle is 40 cm and its length is 14 cm. Find its breadth. Step 1: Use the formula for the perimeter of a rectangle: P = 2(l+b), where P is the perimeter, l is the length, and b is the breadth. Substitute the given values: P=40 cm and l=14 cm. 40 = 2(14+b) Step 2: Divide both sides by 2. (40)/(2) = 14+b 20 = 14+b Step 3: Subtract 14 from both sides to find b. 20 - 14 = b b = 6 cm The breadth of the rectangle is: 6 cm Problem 10: A bag contains 3 red balls, 4 blue balls and 5 green balls. One ball is drawn at random. What is the probability of getting a blue ball? Step 1: Calculate the total number of balls in the bag. Total balls = 3 (red) + 4 (blue) + 5 (green) = 12 balls Step 2: Identify the number of blue balls. Number of blue balls = 4 Step 3: Calculate the probability of drawing a blue ball using the formula: P(blue ball) = Number of blue ballsTotal number of balls P(blue ball) = (4)/(12) Step 4: Simplify the fraction. P(blue ball) = (1)/(3) The probability of getting a blue ball is: (1)/(3) Problem 11: Simplify: ((2x)/(3))^2. Step 1: Apply the square to both the numerator and the denominator. ((2x)/(3))^2 = ((2x)^2)/(3^2) Step 2: Calculate the squares. = (2^2 x^2)/(3^2) = (4x^2)/(9) The simplified expression is: (4x^2)/(9) Problem 12: Find the value of y if (y)/(2)+3=11. Step 1: Subtract 3 from both sides of the equation. (y)/(2)+3-3 = 11-3 (y)/(2) = 8 Step 2: Multiply both sides by 2. (y)/(2) × 2 = 8 × 2 y = 16 The value of y is: 16 Problem 13: Find the coordinates of the midpoint of the line segment joining (2,3) and (6,7). Step 1: Use the midpoint formula: M = ((x_1+x_2)/(2), (y_1+y_2)/(2)). Let (x_1, y_1) = (2,3) and (x_2, y_2) = (6,7). Step 2: Substitute the coordinates into the formula. M = ((2+6)/(2), (3+7)/(2)) Step 3: Calculate the sums and divide by 2. M = ((8)/(2), (10)/(2)) M = (4,5) The coordinates of the midpoint are: (4,5) Problem 14: In the figure, lines l and m are parallel. Find the value of x. Step 1: Identify the relationship between the angles. The angle 120^ and the angle adjacent to it on line l form a linear pair, so their sum is 180^. Let this adjacent angle be . + 120^ = 180^ = 180^ - 120^ = 60^ Step 2: Recognize that and x^ are corresponding angles because lines l and m are parallel. Corresponding angles are equal. x^ = x^ = 60^ The value of x is: 60 Problem 15: Prove that: (a+b)^2 = a^2+2ab+b^2. Step 1: Start with the left-hand side of the equation. (a+b)^2 Step 2: Expand the square as a product of two binomials. (a+b)^2 = (a+b)(a+b) Step 3: Use the distributive property (FOIL method) to multiply the binomials. (a+b)(a+b) = a · a + a · b + b · a + b · b = a^2 + ab + ba + b^2 Step 4: Combine the like terms (ab and ba are the same). = a^2 + 2ab + b^2 This matches the right-hand side of the equation, thus the identity is proven. (a+b)^2 = a^2+2ab+b^2 All done

Please solve the following problems: 1. If 3x+5=20, find the value of x. 2. Solve: 2(x-3)+5=17. 3. Factorise: x2-5x+6. 4. Expand and simplify: (x+2)(x-3).

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Please solve the following problems: 1. If 3x+5=20, find the value of x. 2. Solve: 2(x-3)+5=17. 3. Factorise: x2-5x+6. 4. Expand and simplify: (x+2)(x-3).

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