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Like terms are combined by adding their coefficients: 5x − 2x gives 3x, and 3y + 7y gives 10y. Terms in different variables cannot be merged into a single term, so the simplified expression is 3x + 10y.
The distributive law requires multiplying each term inside the bracket by 3. So 3 × 2x = 6x and 3 × (−5) = −15, giving 6x − 15.
Squaring comes before multiplying by the coefficient, so (−3)² = 9 first. Then 4 × 9 = 36, and the result is positive because the square removes the negative sign.
Adding 7 to both sides isolates x, giving x = −2 + 7. Since −2 + 7 = 5, the solution is x = 5, which can be checked by substituting back.
The fixed fee is charged once regardless of distance, so it is the constant term. The per-kilometre charge multiplies the number of kilometres, giving 2.5k, and the total is 4 + 2.5k.
A linear function has the form y = mx + b, and its graph is always a straight line with constant gradient m. Curves such as parabolas and exponential curves change gradient, so they are not linear.
Multiply the numerical coefficients: 2 × 3 = 6. Then apply the index law for products, adding the powers of the same base: x¹ × x² = x³. The result is 6x³.
Subtract 2l from both sides to get P − 2l = 2w, then divide both sides by 2. This isolates w and gives w = (P − 2l)/2.
Expand the bracket to get 3x + 6 = 21, then subtract 6 from both sides to obtain 3x = 15. Dividing both sides by 3 gives x = 5, which checks correctly.
Using the distributive law, x(x − 3) = x² − 3x and 4(x − 3) = 4x − 12. Combining the like terms −3x + 4x gives +x, so the result is x² + x − 12.
The highest common factor of 6x² and 9x is 3x. Dividing each term by 3x gives 2x and −3, so the factorised form is 3x(2x − 3).
In the form y = mx + b, the coefficient of x is the gradient and the constant is the y-intercept. Here m = 4 and b = −7, so the gradient is 4 and the y-intercept is −7.
Gradient equals the change in y divided by the change in x. The rise is 13 − 5 = 8 and the run is 6 − 2 = 4, so the gradient is 8 ÷ 4 = 2.
Subtract 5 from both sides to get −2x > 6. Dividing by the negative number −2 reverses the inequality sign, giving x < −3.
Multiply both sides by 2 to get 2A = b + h, then subtract b from both sides. This isolates h and gives h = 2A − b.
The monthly fee of $30 is fixed, so it is the constant term. Each extra megabyte adds $0.10, giving the variable term 0.1d, so the model is C = 30 + 0.1d.
Both equations have the same gradient of 3 but different y-intercepts. Lines with equal gradients are parallel, and because the intercepts differ they never intersect.
Adding the two equations eliminates y, giving 3x = 12, so x = 4. Substituting back into x − y = 1 gives y = 3, and checking in 2x + y confirms 8 + 3 = 11.
Divide the coefficients to get 2/4 = 1/2, then subtract indices for each variable: x³⁻¹ = x² and y²⁻⁵ = y⁻³. Writing the negative index in the denominator gives x²/(2y³).
Set 24000 − 3000t = 9000, then subtract 24000 from both sides to get −3000t = −15000. Dividing by −3000 gives t = 5, meaning the value falls by $15 000 over 5 years at $3000 per year.
Two numbers multiplying to −24 and adding to −5 are −8 and 3, so the equation factorises as (x − 8)(x + 3) = 0. Setting each factor to zero gives x = 8 or x = −3.
A point is inside the region when it satisfies both inequalities. For (1, 4): 4 ≤ 2(1) + 3 = 5 is true and 4 ≥ 0 is true, so the point lies inside.
Perpendicular gradients multiply to −1, so the required gradient is −1/2. Using y = mx + b with the point (1, 6): 6 = −0.5(1) + b gives b = 6.5, so y = −0.5x + 6.5.
Only the constant term changes, from 1 to 6, while the x² and x terms stay the same. Adding a constant to a function shifts its graph vertically upwards by that amount, so the graph moves 5 units up.
Set revenue equal to cost: 15n = 8n + 210, so 7n = 210 and n = 30. The difference between revenue and cost per item is 15 − 8 = $7, which is the contribution each sale makes towards covering the fixed cost.
Substitute the given values into the expression and apply the order of operations: 3(4) + 2(−5) = 12 − 10 = 2. Multiplication is performed before addition, and the negative sign must be carried into the second term.
Expand the bracket first: 5(2x − 3) = 10x − 15. Then collect like terms by subtracting 4x, giving 10x − 4x − 15 = 6x − 15.
Subtract 3 from both sides to obtain x/4 = 4, then multiply both sides by 4 to get x = 16. Each step must be applied to both sides to keep the equation balanced.
The fixed flagfall of $4 is paid once, so it is a constant term. The distance charge is $2 for every kilometre, giving the variable term 2k, and the total cost is the sum 4 + 2k.
In the form y = mx + b, the value b is the y-intercept, the y-coordinate where the line crosses the y-axis. Only y = 5x − 3 has b = −3, so this line passes through (0, −3).
All three terms are like terms because they share the same variable m. Combine the coefficients: 7 − 2 + 3 = 8, so the simplified expression is 8m.
A vertical translation down by 3 units changes only the y-intercept, subtracting 3 from the constant term. The gradient stays the same, so y = 2x + 1 − 3 becomes y = 2x − 2.
Add x to both sides to gather the variable terms on one side: 9 = 3x. Dividing both sides by 3 gives x = 3, which checks correctly in the original equation.
Apply the distributive law to each pair of terms: 2x·x = 2x², 2x·5 = 10x, (−1)·x = −x and (−1)·5 = −5. Collecting the middle terms gives 10x − x = 9x, so the result is 2x² + 9x − 5.
Two numbers are needed whose product is −15 and whose sum is +2. The pair 5 and −3 satisfies both conditions, so the factorised form is (x + 5)(x − 3).
The variable r is multiplied by 2π, so undo this by dividing both sides by 2π. This gives r = C/(2π), isolating r on one side of the equation.
The gradient is (8 − 2)/(3 − 1) = 6/2 = 3. Substituting the point (1, 2) into y = 3x + b gives 2 = 3 + b, so b = −1 and the line is y = 3x − 1.
Substitute x = y + 2 into the first equation: 3(y + 2) + 2y = 16, which simplifies to 5y + 6 = 16, so y = 2. Then x = y + 2 = 4, and both values satisfy the original equations.
Subtract 2x from both sides to get 2x − 3 ≤ 7, then add 3 to get 2x ≤ 10. Dividing by the positive number 2 leaves the inequality sign unchanged, so x ≤ 5.
Set the model equal to the required value: 18000 − 1500t = 9000. Subtracting 18000 from both sides gives −1500t = −9000, and dividing by −1500 gives t = 6.
Divide the coefficients: 6 ÷ 2 = 3. For the variables, subtract the index of the denominator from that of the numerator, giving x⁴⁻¹ = x³ and y¹⁻³ = y⁻², so the result is 3x³/y².
The gradient of the given line is (2 − 0)/(4 − 0) = 1/2. Perpendicular gradients multiply to −1, so the required gradient is the negative reciprocal of 1/2, which is −2.
The x-intercepts are found by factorising: x² − 6x + 5 = (x − 1)(x − 5), giving x = 1 and x = 5. The axis of symmetry lies midway between the intercepts, at x = (1 + 5)/2 = 3, which also equals −b/(2a).
Expanding 2(x − 4) gives 2x − 8, so the left side becomes 3x + 2x − 8 = 5x − 8. The student correctly combined the x terms but failed to multiply −4 by 2.
Parallel lines have equal gradients, and the gradient is the coefficient of x in y = mx + b. The second line has gradient 4, so the first line must also have k = 4.
The length is w + 3, so the perimeter is 2w + 2(w + 3) = 4w + 6. Setting 4w + 6 = 34 gives 4w = 28 and w = 7 cm, with length 10 cm.
A quadratic has exactly one solution when its discriminant b² − 4ac equals zero. Here a = 1 and c = 9, so b² − 36 = 0, giving b² = 36 and b = ±6.
The maximum y occurs where the two boundary lines meet: x − 1 = −2x + 8 gives 3x = 9, so x = 3 and y = 3 − 1 = 2. Above this intersection the two inequalities cannot both hold, so the largest y is 2.
Substitute x = −2 carefully: 2(−2)² = 2(4) = 8, then −3(−2) = +6, and finally +1. Adding gives 8 + 6 + 1 = 15, so f(−2) = 15.
Replacing x with (x − 3) shifts the graph 3 units in the positive x-direction, because the vertex moves to x = 3. Adding 4 outside the square shifts the graph 4 units up, so the vertex moves from (0, 0) to (3, 4).