Linear / Rational / Variable Equations - PSAT Math

To solve, remove the radical by squaring both sides

(√3x) 2 = 92

3x = 81

x = 81/3 = 27

First, simplify the equation:

√(8y) + 18 = 4

√(8y) = -14

Then square both sides

(√8y) 2 = -142

8y = 196

y = 196/8 = 24.5

1. Substitute y in the equation for 4.
2. You now have 6 * 4 = 10z + 4
3. Simplify the equation: 24 = 10z + 4
4. Subtract 4 from both sides: 24 – 4 = 10z + 4 – 4
5. You now have 20 = 10z
6. Divde both sides by 10 to solve for z.
7. z = 2.

This goes up at a constant number between values, making it an arthmetic sequence. The first number is 2, with a difference of 3. Plugging this into the arithmetic equation you get An = 2 + 3 (n – 1). Plugging in 20 for n, you get a value of 59.

Looking at the sequence you can see that it doubles each term, making it a geometric sequence. Since it doubles r = 2 and the first term is 5. Plugging this into the geometric equation you get An = 5(2)n–1. Setting n = 6, you get 160 as the 6th term.

The zeroes of the equation are where f(x) = 0 (aka x-intercepts). Setting the equation equal to zero you get x2 = 9. Since a square makes a negative number positive, x can be equal to 3 or –3.

In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.

Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.

A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.

Multiply the equation on both sides by LCM :

or

Substitution confirms that these are the solutions.

There are two solutions of unlike sign.

Multiply both sides by LCD :

or

There are two solutions of unlike sign.

Since all of the equations have the same symbols save for one number, the problem is essentially as follows:

For what value of does the equation

have a solution set other than the empty set?

We can simplify as follows:

If and are not equivalent expressions, the solution set is the empty set. If and are equivalent expressions, the solution set is the set of all real numbers; this happens if and only if:

Therefore, the only equation among the given choices whose solution set is not the empty set is the equation

which is the correct choice.

The problem is basically asking for what value of the equation

has no solution.

We can simplify as folllows:

Since the absolute value of a number must be nonnegative, regardless of the value of , this equation can never have a solution. Therefore, the correct response is that none of the given equations has a solution.

We can examine each individually.

This equation has a solution.

This equation has a solution.

This equation has a solution.

This equation has no solution, since a fourth root of a number must be nonnegative.

The correct choice is .

By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.