SAT Math: Algebra - Linear Equations, Functions, and Systems

Comprehensive practice flashcards covering SAT Algebra questions on linear equations in one and two variables, systems of linear equations, linear functions, and inequalities.

In the equation $\frac{12x+28}{4} - \frac{s}{13} = rx - 8r$, where $r$ and $s$ are constants, what is the value of $s$ if the equation has infinitely many solutions?

$403$. Rewriting the equation yields $3x + 7 - \frac{s}{13} = rx - 8r$. For infinitely many solutions, coefficients of $x$ must match ($r = 3$) and constants must match ($7 - \frac{s}{13} = -8(3) = -24$), which gives $31 = \frac{s}{13}$, resulting in $s = 403$.

What system of linear equations is represented by the two intersecting lines shown in the graph?

The system is $\begin{cases} 4x + 5y = 32 \\ 4x + 10y = 32 \end{cases}$. One line passes through $(8, 0)$ and $(3, 4)$ with slope $m = -\frac{4}{5}$, giving $4x + 5y = 32$. The other line passes through $(8, 0)$ and $(3, 2)$ with slope $m = -\frac{2}{5}$, giving $2x + 5y = 16$, which is equivalent to $4x + 10y = 32$.

Line $k$ is defined by $y = -\frac{17}{3}x + 5$. Line $j$ is perpendicular to line $k$ in the xy-plane. What is the slope of line $j$?

$\frac{3}{17}$ (or approximately $0.1764$ / $0.1765$). The slope of a perpendicular line is the negative reciprocal of the given slope: $\frac{-1}{-17/3} = \frac{3}{17}$.

A cargo helicopter delivers $100$-pound packages and $120$-pound packages. For each trip, it must carry at least $10$ packages and the total weight can be at most $1{,}100$ pounds. What is the maximum number of $120$-pound packages it can carry per trip?

$5$. Let $a$ be the number of $120$-pound packages and $b$ be the number of $100$-pound packages. The system is $120a + 100b \le 1{,}100$ and $a + b \ge 10$. Minimizing $b$ using $b = 10 - a$ gives $120a + 100(10 - a) \le 1{,}100$, which simplifies to $20a \le 100$, or $a \le 5$.

The table gives coordinates of two points on a line in the xy-plane: $(k, 13)$ and $(k+7, -15)$. If the y-intercept of the line is $(k-5, b)$, where $k$ and $b$ are constants, what is the value of $b$?

$33$. The slope of the line is $m = \frac{-15 - 13}{(k+7) - k} = \frac{-28}{7} = -4$. Using the point $(k, 13)$ and the y-intercept $(k-5, b)$, the slope is $\frac{13 - b}{k - (k-5)} = \frac{13 - b}{5} = -4$. Solving yields $13 - b = -20$, so $b = 33$.

Under what graphical and algebraic conditions does a system of two linear equations in two variables have no solution?

The lines in the xy-plane must be parallel and distinct, which algebraically means their slopes are equal ($m_1 = m_2$) while their y-intercepts are different ($b_1 \ne b_2$).

A storage bin contained $24{,}000$ bushels of corn when Hector began using an auger to remove corn at a constant rate. After $5$ hours, $19{,}350$ bushels remained. At this rate, how many total hours will Hector have been using the auger when $12{,}840$ bushels remain?

$12\,\text{hours}$. The rate of removal is $\frac{24{,}000 - 19{,}350}{5} = 930\,\text{bushels per hour}$. The equation is $24{,}000 - 930x = 12{,}840$, which rearranges to $930x = 11{,}160$, giving $x = 12$.

An economist modeled demand $Q$ as a linear function of selling price $P$. Demand was $20{,}000$ units at $P = \$40$ and $15{,}000$ units at $P = \$60$. Based on this model, what is the demand when the selling price is $\$55$ per unit?

$16{,}250\,\text{units}$. The slope is $m = \frac{15{,}000 - 20{,}000}{60 - 40} = -250\,\text{units per dollar}$. The model is $Q(P) = -250(P - 40) + 20{,}000 = -250P + 30{,}000$. At $P = 55$, $Q(55) = -250(55) + 30{,}000 = 16{,}250$.

An apprentice enrolled in training courses modeled by the equation $15x + 20y = 105$, where $x$ is the number of on-site courses and $y$ is the number of online courses. How many more hours does each online training course take than each on-site training course?

$5\,\text{hours}$. In the equation, $15$ represents the number of hours each on-site course takes, and $20$ represents the number of hours each online course takes. The difference is $20 - 15 = 5\,\text{hours}$.

Hiro purchased $4$ shirts and $2$ pairs of pants for $\$86$, and Sofia purchased $3$ shirts and $5$ pairs of pants for $\$166$. What system of linear equations represents this situation if $x$ is the price of each shirt and $y$ is the price of each pair of pants?

The system is $\begin{cases} 4x + 2y = 86 \\ 3x + 5y = 166 \end{cases}$. Hiro's total spent is represented by $4x + 2y = 86$, and Sofia's total spent is represented by $3x + 5y = 166$.

A window repair specialist charges $\$80$ for the first two hours of repair plus an hourly fee for each additional hour. If the total cost for $14$ hours of repair is $\$296$, what function gives the total cost $C(t)$ in dollars for $t$ hours of repair, where $t \ge 2$?

$C(t) = 18t + 44$ (or $C(t) = 80 + 18(t - 2)$). The additional $12$ hours cost $\$296 - \$80 = \$216$, yielding an hourly rate of $\frac{216}{12} = \$18\,\text{per hour}$. Expanding $C(t) = 80 + 18(t - 2)$ gives $C(t) = 18t + 44$.

According to a model, the head width (in millimeters) of a worker bumblebee can be estimated by adding $0.6$ to four times its body weight (in grams). What is the estimated head width of a bumblebee with a body weight of $0.5\,\text{grams}$?

$2.6\,\text{mm}$. The model is represented by $y = 4x + 0.6$. Substituting $x = 0.5$ yields $y = 4(0.5) + 0.6 = 2.0 + 0.6 = 2.6\,\text{mm}$.

The front of a roller-coaster car is initially $15\,\text{feet}$ above the ground and rises at a constant rate of $8\,\text{feet per second}$. What linear equation gives the height $h$ (in feet) of the front of the car $s$ seconds after it starts up the hill?

$h = 8s + 15$. The initial height of $15\,\text{feet}$ corresponds to the constant term $15$, and the ascent rate of $8\,\text{feet per second}$ is represented by the variable term $8s$.

For a system of linear equations written in standard form, $A_1 x + B_1 y = C_1$ and $A_2 x + B_2 y = C_2$, what ratio condition indicates that the system has no solution?

The coefficients must satisfy $\frac{A_1}{A_2} = \frac{B_1}{B_2} \ne \frac{C_1}{C_2}$. This ensures that the lines have the same slope (parallel) but different y-intercepts (distinct).