ACT · Math · Preparing for Higher Math
Ecuaciones e inecuaciones lineales, sistemas, cuadráticas y valor absoluto.
Ecuaciones, inecuaciones, sistemas y expresiones.
$-3x+7>19$ → $-3x>12$ → $x<-4$.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
What value of $x$ satisfies the equation $5(x-2)=3x+8$ ?
Respuesta correcta: C. $9$
Distribute: $5x-10=3x+8$. Subtract $3x$ and add 10: $2x=18$, so $x=9$. Check: $5(9-2)=35$ and $3(9)+8=35$.
5 comes from not distributing ($5x-2=3x+8$). 18 is the value of $2x$, not $x$. $-1$ comes from moving $-10$ to the other side with the wrong sign ($2x=-2$).
What is the sum of all solutions to the equation $|2x-5|=9$ ?
Respuesta correcta: B. $5$
$|2x-5|=9$ means $2x-5=9$ or $2x-5=-9$. The first gives $x=7$ and the second gives $x=-2$. The sum is $7+(-2)=5$.
$-2$ and 7 are each only one of the two solutions. 9 is the value on the right side of the equation, not a sum of solutions.
What is the set of all real solutions of the equation below?
$$\frac{x}{x-2}+\frac{2}{x+2}=\frac{8}{x2-4}$$
Respuesta correcta: A. $\{-6\}$
Multiply both sides by $x2-4=(x-2)(x+2)$: $x(x+2)+2(x-2)=8$, so $x2+4x-12=0$ and $(x+6)(x-2)=0$. But $x=2$ makes the denominators $x-2$ and $x2-4$ equal to 0, so it is extraneous. Check $x=-6$: $\frac{-6}{-8}+\frac{2}{-4}=\frac34-\frac12=\frac14$ and $\frac{8}{32}=\frac14$.
$\{-6, 2\}$ keeps the extraneous root (the classic trap). $\{2\}$ keeps only the extraneous root. "No real solution" throws away $-6$ as well.