SAT Math - Solving Equations

Explanation

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

Add on both sides. Since is greater than and is negative, our answer is negative. We treat as a normal subtraction.

Divide on both sides. When dividing with a negative number, our answer is negative.

Solve for .

Explanation

This is a one step, one variable problem. This means we want to isolate x on one side of the equation with all other constants on the other side. To do this perform the opposite operation to manipulate the equation

Add on both sides.

Solve for the unknown variable:

$\frac{12}{7}$

$\frac{1}{6}$

$\frac{7}{12}$

Explanation

To solve , group like terms on one side of the equal sign.

$x=\frac{12}{7}$

Solve for .

$\pm29$

$\pm39$

Explanation

Take the square root on both sides. We also need to consider having a negative answer.

Remember, two negatives multiplied equals a positive number.

$x=\pm29$

Solve for .

Explanation

Add to both sides.

Solve the equation:

$-\frac{10}{3}$

$-\frac{1}{6}$

$-\frac{1}{3}$

$\frac{1}{3}$

$-\frac{20}{3}$

Explanation

Add nine on both sides.

Divide by negative six on both sides.

$\frac{-6x }{-6}=\frac{ 20}{-6}$

The answer is: $-\frac{10}{3}$

Solve for .

$\frac{3}{4}(x-8)=36$

Explanation

There are TWO ways:

Method : (not really preferred)

$\frac{3}{4}(x-8)=36$ Distribute $\frac{3}{4}$ to each term in the parantheses.

$\frac{3x}{4}-6=36$ Add to both sides.

$\frac{3x}{4}=42$ Multiply by the reciprocal $\frac{4}{3}$ on both sides.

Method : (preferred)

$\frac{3}{4}(x-8)=36$ Multiply by the reciprocal $\frac{4}{3}$ on both sides.

Add to both sides.

Define a function $f(x) =5x- \cos 2x$ .

for exactly one value of on the interval . Which of the following is true of ?

$c\in (1.4, 1.5)$

$c \in (1.0, 1.1 )$

$c \in (1.1, 1.2)$

$c \in (1.2, 1.3 )$

$c \in (1.3, 1.4)$

Explanation

Define $g(x)= f(x) - 8 =5x- \cos 2x - 8$ . Then, if , it follows that .

By the Intermediate Value Theorem (IVT), if is a continuous function, and and are of unlike sign, then for some $c \in (a, b)$ . and $\cos 2x$ are both continuous everywhere, so is a continuous function, so the IVT applies here.

Evaluate for each of the following values: $\left { 1.0, 1.1, 1.2, 1.3, 1.4, 1.5\right }$

$g(1) =5 (1)- \cos 2 (1)- 8$

$=5 - \cos 2 - 8$

$\approx 5 - (-0.42) - 8$

$\approx -2.58$

$g(1.1) =5 (1.1)- \cos 2 (1.1)- 8$

$=5.5 - \cos 2.2 - 8$

$\approx 5.5 - (-0.59) - 8$

$\approx -1.91$

$g(1.2) =5 (1.2)- \cos 2 (1.2)- 8$

$=6 - \cos 2.4 - 8$

$\approx 6 - (-0.74) - 8$

$\approx -1.26$

$g(1.3) =5 (1.3)- \cos 2 (1.3)- 8$

$=6.5 - \cos 2.6 - 8$

$\approx 6.5 -(-0.86) - 8$

$\approx -0.64$

$g(1.4) =5 (1.4)- \cos 2 (1.4)- 8$

$=7 - \cos 2.8 - 8$

$\approx 7 - (-0.94) - 8$

$\approx -0.06$

$g(1.5) =5 (1.5)- \cos 2 (1.5)- 8$

$=7.5 - \cos 3 - 8$

$\approx 7.5 -(-0.99)- 8$

$\approx 0.49$

Only in the case of does it hold that assumes a different sign at both endpoints - . By the IVT, , and , for some $c\in (1.4, 1.5)$ .

Solve for .

Explanation

To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.