Linear growth and decay

Given \(g(x)=60-4x\), \(g(3)=60-4\times3=48\) and \(g(8)=60-4\times8=28\). Therefore, \(g(3)+g(8)=48+28=76\). Option 80 can result from an error in multiplication or subtraction. Exam tip: substitute each input into the function separately before adding the results.

For the two expressions to be equal, set \(10+4x=20+2x\). Subtracting \(2x\) from both sides gives \(10+2x=20\), so \(2x=10\) and \(x=5\). Checking: \(A(5)=30\) and \(B(5)=30\). Option \(x=4\) is a close distractor, but it gives 26 and 28, which are not equal. Exam tip: To find where two linear expressions are equal, equate them first and solve for \(x\).

For the two expressions to be equal, set \(C(x)=D(x)\): \(100-5x=80-x\). This gives \(20=4x\), so \(x=5\). Therefore, option B is correct. Substituting \(x=4\) does not give equal values, so that close distractor is incorrect. Exam tip: To find where two linear expressions are equal, equate them first and then solve for \(x\).

Given y=15+5x and y=55, substitute 55 for y: 15+5x=55. Thus, 5x=40 and x=8. Therefore, 8 is the correct option. For example, if x=7, then y=15+35=50, not 55. Exam tip: In a linear expression, substitute the given value of y and isolate the term containing x first.

Given (y=96-8x) and (y=40), we get (96-8x=40). Subtracting 96 from both sides gives (-8x=-56), so (x=7). If x=6, then y=48, not 40. Exam tip: When moving or simplifying negative terms, track the signs on both sides carefully.

Here, the increase in x is 11 - 1 = 10. In y = 2x + 30, the coefficient of x is 2, so the change in y is 2 × 10 = 20. Therefore, y increases by 20. The constant term 30 changes the starting value of y, not the amount of increase. Exam tip: In a linear equation y = mx + c, multiply the change in x by m to find the change in y.

As x changes from 3 to 13, the change in x is 10. In y = 150 - 6x, the coefficient of x is -6, so y decreases by 6 for every increase of 1 in x. Therefore, the total change in y is -6 × 10 = -60; hence, y decreases by 60. Checking directly, y(3) = 132 and y(13) = 72. The value 78 is not the final value of y, and y does not increase. Exam tip: In a linear equation, total change = coefficient of x × change in x.

For linear growth, the coefficient of (x) must be positive and nonzero. (y=18+2x) is correct.

In linear decay, the power of \(x\) is 1 and its coefficient is negative. In \(y=45-3x\), the coefficient of \(x\) is \(-3\), so \(y\) decreases by 3 for every one-unit increase in \(x\). \(y=45+3x\) shows linear growth, while \(y=45-3x^2\) is quadratic rather than linear. Exam tip: for \(y=a+bx\), check that \(b<0\) to identify linear decay.

Given F(t)=18+rt and F(2)=30, we get 18+2r=30. Thus, 2r=12 and r=6. Now, F(5)=18+6×5=48. Therefore, 48 is the correct answer. The value 42 would result from using t=4, so it is not correct here. Exam tip: first find r from the given function value, then substitute the required value of t.

Given \(H(t)=24+5t\). Substituting \(t=4\), \(H(4)=24+5(4)=24+20=44\) cm. Therefore, 44 cm is correct. The value 49 cm would result from an error in multiplication or addition. Exam tip: To evaluate a linear expression, substitute the given value for the variable and calculate step by step.

For a linear polynomial \(ax+b\), the sign of coefficient \(a\) determines its direction of change. In \(-3x+7\), the coefficient is \(-3\), so its value decreases as \(x\) increases. A positive coefficient, as in \(5x+2\), indicates growth. Exam tip: check the coefficient of \(x\) first.

In \(75-4x\), the coefficient of x is \(-4\), so the quantity falls by 4 units for every increase of 1 in x. \(75+4x\) shows linear growth. Exam tip: for linear decay, look for a negative coefficient of x.

In \(R(t)=90-5t\), the coefficient of \(t\) is \(-5\), so \(R\) falls by 5 units when \(t\) rises by 1. \(90+5t\) represents growth. Exam tip: a negative linear coefficient indicates decay.

For the amount to become 620 rupees, form the equation 300+40m=620. Thus, 40m=320, so m=8. Therefore, the amount will be 620 rupees after 8 months. With 7 months, the amount would be only 580 rupees, so it is not correct. Exam tip: Substitute the target value in a linear expression and solve for the variable.

For the battery percentage to be 48, set \(B(h)=48\): \(96-6h=48\). Thus, \(6h=48\), so \(h=8\). Therefore, the battery will be 48% after 8 hours. At 6 hours, it is \(96-36=60\)% remaining, so that nearby option is incorrect. Exam tip: In a linear decay problem, equate the function to the target value and solve for the variable.

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the change per step. Here, \(a=28\) and the growth is \(b=9\), so the rule is \(y=28+9x\). In \(y=28-9x\), the value decreases by 9 each step, so it represents decay, not growth. Exam tip: put \(x=0\); the correct rule must give \(y=28\).

The initial value is the constant term, so it is 180. A decrease of 15 per step means that the coefficient of \(x\) must be \(-15\). Therefore, the rule is \(y=180-15x\). In option A, the value increases rather than decreases. Exam tip: in a linear rule \(y=a+bx\), a negative \(b\) indicates decrease.

In \(y=18-3x\), the coefficient of x is \(-3\). Thus, for every increase of 1 in x, y decreases by a constant 3, which is linear decay. \(18+3x\) shows linear growth instead. Exam tip: look for a negative coefficient of x in a linear relation.

The change in \(x\) is \(11-4=7\). The coefficient of \(x\) is \(-5\), so for every increase of 1 in \(x\), \(y\) decreases by 5. Hence, the total decrease in \(y\) is \(5\times 7=35\). Equivalently, \(y(4)=75\) and \(y(11)=40\), so the decrease is \(75-40=35\). Exam tip: multiply the magnitude of the coefficient by the change in the variable to find total decrease.