Understand and apply concepts of equations

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HiSET › Understand and apply concepts of equations

Questions 1 - 10

Solve for :

$\frac{y+1}{x+1} = 3$

$x = \frac{1}{3} y - \frac{2}{3}$

$x = \frac{1}{3} y + \frac{4}{3}$

$x = \frac{2}{3} y -\frac{4}{3}$

$x = \frac{2}{3} y + \frac{1}{3}$

$x = \frac{4}{3} y + \frac{1}{3}$

Explanation

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, take the reciprocal of both sides:

$\frac{y+1}{x+1} = 3$

$\frac{x+1}{y+1} = \frac{1}{3}$

Multiply both sides by :

$\frac{x+1}{y+1} \cdot (y+1)= \frac{1}{3}(y+1)$

$x+1= \frac{1}{3}(y+1)$

Distribute on the right:

$x+1= \frac{1}{3} \cdot y+ \frac{1}{3} \cdot 1$

$x+1= \frac{1}{3} y+ \frac{1}{3}$

Subtract 1 from both sides, rewriting 1 as $\frac{3}{3}$ to facilitate subtraction:

$x+1 - 1 = \frac{1}{3} y+ \frac{1}{3} - 1$

$x = \frac{1}{3} y+ \frac{1}{3} - \frac{3}{3}$

$x = \frac{1}{3} y - \frac{2}{3}$ ,

the correct response.

Solve the following equation:

$\sqrt[3]{5-2x}+5=4$

$x=\sqrt[3]{5}$

Explanation

$\sqrt[3]{5-2x}+5=4$

The first step to solving an equation where is in a radical is to isolate the radical. To do this, we need to subtract the 5 from both sides.

$\sqrt[3]{5-2x}+5-5=4-5$

$\sqrt[3]{5-2x}=-1$

Now that the radical is isolated, clear the radical by raising both sides to the power of 3. Note:

$(-1)^3 = -1 \cdot -1 \cdot -1 = -1$

$(\sqrt[3]{5-2x})^3=(-1)^3$

Now we want to isolate the term. First, subtract the 5 from both sides.

Finally, divide both sides by to solve for .

$\frac{-2x}{-2}=\frac{-6}{-2}$

Solve for :

$\frac{y+1}{x+1} = 3$

$x = \frac{1}{3} y - \frac{2}{3}$

$x = \frac{1}{3} y + \frac{4}{3}$

$x = \frac{2}{3} y -\frac{4}{3}$

$x = \frac{2}{3} y + \frac{1}{3}$

$x = \frac{4}{3} y + \frac{1}{3}$

Explanation

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, take the reciprocal of both sides:

$\frac{y+1}{x+1} = 3$

$\frac{x+1}{y+1} = \frac{1}{3}$

Multiply both sides by :

$\frac{x+1}{y+1} \cdot (y+1)= \frac{1}{3}(y+1)$

$x+1= \frac{1}{3}(y+1)$

Distribute on the right:

$x+1= \frac{1}{3} \cdot y+ \frac{1}{3} \cdot 1$

$x+1= \frac{1}{3} y+ \frac{1}{3}$

Subtract 1 from both sides, rewriting 1 as $\frac{3}{3}$ to facilitate subtraction:

$x+1 - 1 = \frac{1}{3} y+ \frac{1}{3} - 1$

$x = \frac{1}{3} y+ \frac{1}{3} - \frac{3}{3}$

$x = \frac{1}{3} y - \frac{2}{3}$ ,

the correct response.

What is 25% of ?

Explanation

Solve for in the equation

by isolating on the left side. Do this by reversing the operations in the reverse of the order of operations.

First, subtract 17 from both sides:

Now, divide both sides by 2:

$2x \div 2 = 82 \div 2$

One way to find 25% of this value is to multiply 41 by 25 and divide by 100:

$\frac{41 \times 25}{100} = \frac{1,025}{100} = 10.25$ ,

the correct choice.

What is the vertex of the following quadratic polynomial?

$(\frac{3}{4},\frac{127}{8})$

$(-\frac{5}{3},\frac{12}{7})$

$(\frac{4}{3},\frac{98}{17})$

$(-\frac{4}{9},-\frac{58}{17})$

$(-\frac{4}{23},-\frac{15}{17})$

Explanation

Given a quadratic function

the vertex will always be

$(-\frac{b}{2a},\frac{4ac-b^2}{4a})$ .

Thus, since our function is

, , and .

We plug these variables into the formula to get the vertex as

$(-\frac{-3}{2\times2},\frac{4\times2\times17-(-3)^2}{4\times2})$

$(-\frac{-3}{4},\frac{136-9}{8})$

$=(\frac{3}{4},\frac{127}{8})$ .

Hence, the vertex of

$(\frac{3}{4},\frac{127}{8})$ .

What is 25% of ?

Explanation

Solve for in the equation

by isolating on the left side. Do this by reversing the operations in the reverse of the order of operations.

First, subtract 17 from both sides:

Now, divide both sides by 2:

$2x \div 2 = 82 \div 2$

One way to find 25% of this value is to multiply 41 by 25 and divide by 100:

$\frac{41 \times 25}{100} = \frac{1,025}{100} = 10.25$ ,

the correct choice.

What is the vertex of the following quadratic polynomial?

$(\frac{3}{4},\frac{127}{8})$

$(-\frac{5}{3},\frac{12}{7})$

$(\frac{4}{3},\frac{98}{17})$

$(-\frac{4}{9},-\frac{58}{17})$

$(-\frac{4}{23},-\frac{15}{17})$

Explanation

Given a quadratic function

the vertex will always be

$(-\frac{b}{2a},\frac{4ac-b^2}{4a})$ .

Thus, since our function is

, , and .

We plug these variables into the formula to get the vertex as

$(-\frac{-3}{2\times2},\frac{4\times2\times17-(-3)^2}{4\times2})$

$(-\frac{-3}{4},\frac{136-9}{8})$

$=(\frac{3}{4},\frac{127}{8})$ .

Hence, the vertex of

$(\frac{3}{4},\frac{127}{8})$ .

Solve the following equation:

$\sqrt[3]{5-2x}+5=4$

$x=\sqrt[3]{5}$

Explanation

$\sqrt[3]{5-2x}+5=4$

The first step to solving an equation where is in a radical is to isolate the radical. To do this, we need to subtract the 5 from both sides.

$\sqrt[3]{5-2x}+5-5=4-5$

$\sqrt[3]{5-2x}=-1$

Now that the radical is isolated, clear the radical by raising both sides to the power of 3. Note:

$(-1)^3 = -1 \cdot -1 \cdot -1 = -1$

$(\sqrt[3]{5-2x})^3=(-1)^3$

Now we want to isolate the term. First, subtract the 5 from both sides.

Finally, divide both sides by to solve for .

$\frac{-2x}{-2}=\frac{-6}{-2}$

Which of the following expressions represents the discriminant of the following polynomial?

$9-4\times6\times7$

$6-3\times4\times7$

$9-4\times3\times7$

$7-4\times6\times3$

$6-4\times3\times7+ 0.0028917$

Explanation

The discriminant of a quadratic polynomial

is given by

Thus, since our quadratic polynomial is

, , and .

Plugging these values into the discriminant equation, we find that the discriminant is

$9-4\times6\times7$ .

Which of the following expressions represents the discriminant of the following polynomial?

$9-4\times6\times7$

$6-3\times4\times7$

$9-4\times3\times7$

$7-4\times6\times3$

$6-4\times3\times7+ 0.0028917$

Explanation

The discriminant of a quadratic polynomial

is given by

Thus, since our quadratic polynomial is

, , and .

Plugging these values into the discriminant equation, we find that the discriminant is

$9-4\times6\times7$ .

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