Systems of two linear equations in two variables SAT question

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 $-2$ yields $-2ty=-1-4x$, or $-1-4x=-2ty$. Adding this equation to the first equation in the given system, $4x-6y=10y+2$, yields $(4x-6y)+(-1-4x)=(10y+2)+(-2ty)$, or $-1-6y=10y-2ty+2$. Subtracting $10y$ from both sides of this equation yields $(-1-6y)-(10y)=(10y-2ty+2)-(10y)$, or $-1-16y=-2ty+2$. If the given system has no solution, then the equation $-1-16y=-2ty+2$ has no solution. If this equation has no solution, the coefficients of $y$ on each side of the equation, $-16$ and $-2t$, must be equal, which yields the equation $-16=-2t$. Dividing both sides of this equation by $-2$ yields $8=t$. 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 $Ax+By=C$, 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 $10y$ from both sides of the first equation in the given system yields $(4x-6y)-(10y)=(10y+2)-(10y)$, or $4x-16y=2$. Subtracting $2x$ from both sides of the second equation in the given system yields $(ty)-(2x)=(\frac{1}{2}+2x)-(2x)$, or $-2x+ty=\frac{1}{2}$. 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 $\frac{\frac{1}{2}}{2}$, 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 $-16$ yields $(-\frac{1}{2})(-16)=(-\frac{t}{16})(-16)$, or $t=8$. Therefore, when $t=8$, the lines represented by these equations are parallel. Thus, if the system has no solution, the value of $t$ is $8$.