SAT Question f75bd744: Systems of Linear Equations answer and explanation with Desmos solution

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 $\left(4x-6y\right)+\left(-1-4x\right)=\left(10y+2\right)+\left(-2ty\right)$, or $-1-6y=10y-2ty+2$. Subtracting $10y$ from both sides of this equation yields $\left(-1-6y\right)-\left(10y\right)=\left(10y-2ty+2\right)-\left(10y\right)$, 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 $\left(4x-6y\right)-\left(10y\right)=\left(10y+2\right)-\left(10y\right)$, or $4x-16y=2$. Subtracting $2x$ from both sides of the second equation in the given system yields $\left(ty\right)-\left(2x\right)=\left(\frac{1}{2}+2x\right)-\left(2x\right)$, 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 $\left(-\frac{1}{2}\right)\left(-16\right)=\left(-\frac{t}{16}\right)\left(-16\right)$, 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 $8$.