SAT · Math · Algebra
Sistemas de dos ecuaciones lineales: resolverlos y saber cuándo tienen una, ninguna o infinitas soluciones.
$x+y=10$ y $x-y=2$. Sumando: $2x=12$ → $x=6$ y $y=4$.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
$$\begin{cases} x+y=15 \\ x-y=3 \end{cases}$$
The solution to the given system of equations is $(x, y)$. What is the value of $x$?
Respuesta numérica (escribes el número).
Respuesta: 9
Add the two equations to eliminate $y$: $(x+y)+(x-y)=15+3$, so $2x=18$ and $x=9$.
Check: $y=15-9=6$, and $9-6=3$.
$x+y=14$
$x-y=4$
If $(x, y)$ is the solution to the given system of equations, what is the value of $xy$?
Respuesta correcta: B. $45$
Add the equations: $2x=18$, so $x=9$. Then $y=14-9=5$, and $xy=9\cdot5=45$.
18 is $2x$, not $xy$. 50 and 56 come from pairing the wrong values (for example, $10\cdot5$ or $14\cdot4$).
$ax+4y=10$
$6x+(a+2)y=15$
In the given system of equations, $a$ is a constant. For what value of $a$ does the system have infinitely many solutions?
Respuesta numérica (escribes el número).
Respuesta: 4
Infinitely many solutions means the second equation is a multiple of the first: $\frac{a}{6}=\frac{4}{a+2}=\frac{10}{15}=\frac{2}{3}$. From $\frac{a}{6}=\frac{2}{3}$, $a=4$, and then $\frac{4}{4+2}=\frac{2}{3}$ also holds.
Trap: setting only $\frac{a}{6}=\frac{4}{a+2}$ gives $a2+2a-24=0$, so $a=4$ or $a=-6$. But with $a=-6$ the coefficient ratio is $-1$ while the constant ratio is $\frac{2}{3}$, so the lines are parallel and the system has no solution.