Advanced: Algebra

Advanced Algebra is still linear, and that is the trick. The equations look like the ones from Foundations, but one of the numbers has been replaced by a letter, and the question is no longer "what is x" but "for what value of the constant does this equation have no solution", or "has infinitely many", or "has a solution of x = 5". Nothing new is being solved; the same facts about lines are being asked from the other side.

The facts are few. A linear equation is true for every x exactly when both sides are the same expression, and true for no x exactly when the x-terms match but the constants do not. A system has no solution when the lines are parallel and infinitely many when they are the same line. A linear function is fixed by any two of its values. Every hard Algebra question is one of these facts with a constant to solve for.

This lesson also covers the two-constraint word problems, where a budget and a minimum quantity, or a capacity and a requirement, produce a system of inequalities and the answer is the largest or smallest whole number the constraints allow.

A linear equation in one variable can have exactly one solution, no solution, or infinitely many, and which of the three depends on the coefficients. Bring both sides to the form ax + b = cx + d. If a is not equal to c, subtracting gives (a - c)x = d - b and there is exactly one solution, x = (d - b)/(a - c). If a = c and b = d, the two sides are the same expression and every real number is a solution. If a = c but b is not d, the x-terms cancel to leave a false statement such as 6 = -9 and there is no solution. So 3(x + k) = 3x + 12 has infinitely many solutions when 3k = 12, that is k = 4, and no solution for any other k. The equation kx + 6 = 4x - 9 has no solution when k = 4, because then it reduces to 6 = -9, and exactly one solution for every other k.

A system of two linear equations with an unknown constant is the same question about two lines. Writing the equations as ax + by = c and dx + ey = f, the slopes are equal exactly when a/d = b/e, equivalently ae = bd. If also c/f equals that common ratio, the second equation is a multiple of the first and the system has infinitely many solutions; if not, the lines are parallel and the system has none. For 2x + 5y = 7 and ax + 10y = c, the slopes match when a/2 = 10/5, so a = 4; then the system has infinitely many solutions if c = 14 and no solution for every other c. When a constant appears in a slope-intercept form, such as y = 3x + c against 6x - 2y = 10, substitute: 6x - 2(3x + c) = 10 gives -2c = 10, which is true for every x when c = -5 (infinitely many) and false for every x otherwise (none). A question may equally ask for the value that produces exactly one solution: that is every value except the one that equalizes the slopes.

When an equation contains an unknown constant and you are told a solution, the solution is a number to substitute. If x = 5 solves kx - 3 = 2(x + 4), then 5k - 3 = 18, so k = 21/5. If the line y = mx + 4 passes through (6, 1), then 1 = 6m + 4 and m = -1/2. If the point (a, 2a) lies on 3x - y = 7, then 3a - 2a = 7 and a = 7. The pattern is the same each time: whatever is known goes in, and the equation is solved for the letter that remains.

A linear function f(x) = mx + b is determined by any two of its values, because two points determine a line. Given f(1) = 7 and f(4) = 16, the slope is (16 - 7)/(4 - 1) = 3 and then 7 = 3(1) + b gives b = 4, so f(x) = 3x + 4 and f(10) = 34. A table with two rows, a description like "f increases by 6 whenever x increases by 2", or a graph through two labelled points all reduce to this. Questions then ask for another value, for the intercept, for the x at which f takes a given value, or for a comparison between two linear functions, such as the x at which they are equal.

Parallel lines have equal slopes and perpendicular lines have slopes whose product is -1. To write a line perpendicular to 2x + 3y = 6 through (4, 1): the given slope is -2/3, so the perpendicular slope is 3/2, and point-slope form gives y - 1 = (3/2)(x - 4), which is y = (3/2)x - 5. A line's x-intercept is where y = 0 and its y-intercept where x = 0; a line through (a, 0) and (0, b) with both nonzero can be written x/a + y/b = 1, a form worth recognizing when a question gives intercepts in terms of constants. The perpendicular bisector of a segment passes through its midpoint with the negative reciprocal slope.

Two-constraint word problems produce a system of linear inequalities, and the question asks for the greatest or least value of one quantity. A theater holds at most 300 people, charges 12 dollars for adults and 8 for children, and needs at least 3,000 dollars in ticket sales: with a adults and c children, a + c is at most 300 and 12a + 8c is at least 3,000. The fewest adults occurs when the theater is full of children otherwise, c = 300 - a, giving 12a + 8(300 - a) at least 3,000, so 4a is at least 600 and a is at least 150. The extreme always lies where at least one constraint is an equality, which is why the method is to set one constraint to an equation and substitute it into the other. Check the answer and its neighbor against both original inequalities; when the variable counts people or items, round toward the side that satisfies them.

Solving for x when the question asks for the constant. Read what the question wants before you start moving terms.

Stopping at equal slopes and declaring no solution without checking the intercepts; equal intercepts mean infinitely many.

Matching x-coefficients for "infinitely many" but forgetting the constants must match too.

Taking the reciprocal or the negative for a perpendicular slope, but not both.

Using three points to define a linear function when two are given and the third is what is asked; the slope must come from the two known values.

In a two-constraint problem, minimizing the wrong variable or forgetting that the extreme happens on a boundary.

Rounding a boundary value the wrong way when the variable counts people or boxes: 33.3 boxes means 33 boxes fit, not 34.