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To add numbers with different signs, subtract the smaller absolute value from the larger and keep the sign of the larger absolute value. Here |−7| > |+4|, so the result is −(7 − 4) = −3.
The product of two numbers with the same sign is positive. Multiply the absolute values: 5 × 6 = 30, and since both factors are negative, the result is +30.
The exponent applies to the entire base (−3): (−3) × (−3) = 9. A negative times a negative is positive, so the square of any real number is non-negative.
Factor 50 as 25 × 2, where 25 is a perfect square. Then √50 = √(25 × 2) = √25 × √2 = 5√2.
Simplify each radical first: √18 = √(9×2) = 3√2 and √32 = √(16×2) = 4√2. Adding like radicals gives 3√2 + 4√2 = 7√2.
Combine under one radical: √12 × √27 = √(12 × 27) = √324 = 18. Alternatively, √12 = 2√3 and √27 = 3√3, so 2√3 × 3√3 = 6 × 3 = 18.
Use (x + y)² = x² + 2xy + y² with x = a, y = 3: (a + 3)² = a² + 2·a·3 + 3² = a² + 6a + 9.
This is the difference-of-squares pattern (a + b)(a − b) = a² − b². With a = x and b = 2, the product is x² − 2² = x² − 4.
This is a difference of squares: x² − 16 = x² − 4² = (x − 4)(x + 4).
Find two numbers whose product is 6 and whose sum is 5. The pair 2 and 3 works (2 × 3 = 6, 2 + 3 = 5), so x² + 5x + 6 = (x + 2)(x + 3).
Look for two numbers whose product is +12 and whose sum is −7. A positive product with a negative sum means both numbers must be negative. The pair −3 and −4 works, since (−3)(−4) = 12 and (−3) + (−4) = −7, so x² − 7x + 12 = (x − 3)(x − 4).
Apply the distributive property: 2(x + 5) = 2·x + 2·5 = 2x + 10.
Substitute a = 4 and b = 1: 3(4) − 2(1) = 12 − 2 = 10.
Square both the coefficient and the radical: (2√3)² = 2² × (√3)² = 4 × 3 = 12.
Simplify each radical: √45 = √(9×5) = 3√5 and √20 = √(4×5) = 2√5. Then 3√5 − 2√5 = √5.
This is the difference-of-squares pattern with a = 3 and b = √2: (3 + √2)(3 − √2) = 3² − (√2)² = 9 − 2 = 7.
Use (a + b)² = a² + 2ab + b² with a = √3 and b = √2: (√3)² + 2·√3·√2 + (√2)² = 3 + 2√6 + 2 = 5 + 2√6.
Subtract 3 from both sides: 2x = 8. Then divide by 2: x = 4.
Subtract x from both sides: 2x − 5 = 7. Add 5: 2x = 12. Divide by 2: x = 6.
Subtract 1 from both sides: x/2 = 4. Multiply both sides by 2: x = 8.
Add the two equations: (x + y) + (x − y) = 10 + 4, so 2x = 14 and x = 7. Substitute back: 7 + y = 10, so y = 3.
From the first equation, y = 7 − 2x. Substitute into the second: 3x + 2(7 − 2x) = 12, so 3x + 14 − 4x = 12, giving −x = −2 and x = 2. Then y = 7 − 2(2) = 3.
Factor: x² − 5x + 6 = (x − 2)(x − 3) = 0. So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
Find two numbers whose product is −10 and sum is 3: 5 and −2 work (5·(−2) = −10, 5 + (−2) = 3). So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2.
Factor as (2x − 1)(x − 3) = 0. Check: 2x·x = 2x², 2x·(−3) = −6x, (−1)·x = −x, (−1)(−3) = 3; the x terms combine to −7x. Setting each factor to zero gives 2x = 1 (so x = 1/2) or x = 3.
Add 7 to both sides: x² + 6x = 7. Complete the square by adding (6/2)² = 9: x² + 6x + 9 = 16, i.e. (x + 3)² = 16. Taking square roots gives x + 3 = ±4, so x = 1 or x = −7.
Subtract 1 from both sides: 2x < 6. Divide by 2 (positive, so the inequality keeps its direction): x < 3.
Subtract x from both sides: 2x − 5 ≥ 3. Add 5: 2x ≥ 8. Divide by 2: x ≥ 4.
Divide both sides by −2. Dividing by a negative number flips the inequality sign, so x < −3. Equivalently, x < 6/(−2) = −3.
One pen costs 450 ÷ 3 = ¥150. Seven pens cost 7 × 150 = ¥1050.
Let B's age be b; then A's age is b + 3. The equation is b + (b + 3) = 25, so 2b = 22 and b = 11.
In 2 hours the first train covers 60 × 2 = 120 km and the second covers 80 × 2 = 160 km. Since they travel in opposite directions, the distances add: 120 + 160 = 280 km.
Three apples cost 3 × 80 = ¥240. The remaining cost for two oranges is 420 − 240 = ¥180. One orange costs 180 ÷ 2 = ¥90.
Let the width be w; then the length is w + 3. The area equation is w(w + 3) = 40, i.e. w² + 3w − 40 = 0. Factoring gives (w + 8)(w − 5) = 0, so w = 5 (rejecting the negative root since width must be positive).
In the form y = mx + b, m is the slope. Comparing y = 2x + 3 to y = mx + b gives m = 2.
The slope between (x₁, y₁) and (x₂, y₂) is (y₂ − y₁)/(x₂ − x₁). Here it is (8 − 2)/(3 − 1) = 6/2 = 3.
In the form y = mx + b, the y-intercept is b. Comparing y = 3x − 5 to y = mx + b gives b = −5, the point (0, −5).
The y-intercept is (0, 2), so b = 2. The slope is (8 − 2)/(3 − 0) = 6/3 = 2. Therefore the equation is y = 2x + 2.
Parallel lines have the same slope. The slope of y = 4x − 3 is 4, so any line parallel to it also has slope 4.
The rate of change is the increase in y divided by the increase in x. When x = 1, y = 2(1)² = 2; when x = 3, y = 2(3)² = 18. So the rate of change is (18 − 2)/(3 − 1) = 16/2 = 8. Unlike a linear function, the rate of change of y = ax² depends on which interval of x you choose.
Substitute x = 2 into the relationship y = 6/x, which gives y = 6/2 = 3. This is an inverse proportion, so the product xy stays constant at 6 for every point on the graph. As x grows larger, y becomes correspondingly smaller.
Inverse proportion means xy = k. Using the given point: k = 3 × 4 = 12. When x = 6, y = k/x = 12/6 = 2.
The sign of the leading coefficient a determines the opening direction of y = ax². When a > 0, the parabola opens upward (a smile shape); when a < 0, it opens downward.
Substitute the point into y = ax². With x = 2 and y = 12, the equation becomes 12 = a(2)² = 4a. Dividing both sides by 4 gives a = 3, so the function is y = 3x².
Because the interval −2 ≤ x ≤ 3 contains x = 0, the smallest value of y is y = 0² = 0, not the value at either endpoint. The largest value comes from the endpoint furthest from 0, which is x = 3, giving y = 9. So the range is 0 ≤ y ≤ 9.
Substitute x = 40 into y = 0.01x². First square the speed: 40² = 1600. Then multiply by the constant: y = 0.01 × 1600 = 16 metres. Because braking distance varies with the square of the speed, doubling the speed makes the stopping distance four times longer.
For a linear function the extreme values of y occur at the endpoints of the x-interval. At x = 0, y = 5; at x = 3, y = −2(3) + 5 = −1. Because the coefficient of x is negative, y decreases as x increases, so the range runs from −1 up to 5.
A parabola y = ax² + ... opens downward when the leading coefficient a is negative. Only y = −x² + 2 has a < 0 (a = −1); the others have positive leading coefficients and open upward.
Substitute x = 2: y = (2)² − 3(2) = 4 − 6 = −2.
Replace x with 2x in the function: y = (2x)² = 4x². Comparing this with the original y = x² shows the new value is 4 times the old one. In general, for y = ax² multiplying x by k multiplies y by k².