Question #f75bd744 - SAT Suite Question Bank

ID: f75bd744

Modded SAT Question Bank by Abdullah Mallik

$4 x - 6 y = 10 y + 2$

$t y = \frac{1}{2} + 2 x$

In the given system of equations, $t$ is a constant. If the system has no solution, what is the value of $t$ ?

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Correct Answer: 8

Rationale

The correct answer is $8$ . The given system of equations can be solved using the elimination method. Multiplying both sides of the second equation in the given system by $negative 2$ yields $- 2 t y = - 1 - 4 x$ , or $- 1 - 4 x = - 2 t y$ . Adding this equation to the first equation in the given system, $4 x - 6 y = 10 y + 2$ , yields $(4 x - 6 y) + (- 1 - 4 x) = (10 y + 2) + (- 2 t y)$ , or $- 1 - 6 y = 10 y - 2 t y + 2$ . Subtracting $10 y$ from both sides of this equation yields $(- 1 - 6 y) - (10 y) = (10 y - 2 t y + 2) - (10 y)$ , or $- 1 - 16 y = - 2 t y + 2$ . If the given system has no solution, then the equation $- 1 - 16 y = - 2 t y + 2$ has no solution. If this equation has no solution, the coefficients of $y$ on each side of the equation, $negative 16$ and $minus 2 t$ , must be equal, which yields the equation $- 16 = - 2 t$ . Dividing both sides of this equation by $negative 2$ yields $8 equals t$ . Thus, if the system has no solution, the value of $t$ is $8$ .

Alternate approach: A system of two linear equations in two variables, $x$ and $y$ , has no solution if the lines represented by the equations in the xy-plane are parallel and distinct. Lines represented by equations in the form $A x + B y = C$ , where $A$ , $B$ , and $C$ are constant terms, are parallel if the ratio of the x-coefficients is equal to the ratio of the y-coefficients, and distinct if the ratio of the x-coefficients are not equal to the ratio of the constant terms. Subtracting $10 y$ from both sides of the first equation in the given system yields $(4 x - 6 y) - (10 y) = (10 y + 2) - (10 y)$ , or $4 x - 16 y = 2$ . Subtracting $2 x$ from both sides of the second equation in the given system yields $(t y) - (2 x) = (\frac{1}{2} + 2 x) - (2 x)$ , or $- 2 x + t y = \frac{1}{2}$ . The ratio of the x-coefficients for these equations is $- \frac{2}{4}$ , or $- \frac{1}{2}$ . The ratio of the y-coefficients for these equations is $- \frac{t}{16}$ . The ratio of the constant terms for these equations is $StartFraction one half Over 2 EndFraction$ , or $\frac{1}{4}$ . Since the ratio of the x-coefficients, $- \frac{1}{2}$ , is not equal to the ratio of the constants, $\frac{1}{4}$ , the lines represented by the equations are distinct. Setting the ratio of the x-coefficients equal to the ratio of the y-coefficients yields $- \frac{1}{2} = - \frac{t}{16}$ . Multiplying both sides of this equation by $negative 16$ yields $(- \frac{1}{2}) (- 16) = (- \frac{t}{16}) (- 16)$ , or $t equals 8$ . Therefore, when $t equals 8$ , the lines represented by these equations are parallel. Thus, if the system has no solution, the value of $t$ is $8$ .

Question Difficulty: Hard