Linear / Rational / Variable Equations - PSAT Math

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.

Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.

Start by combining like terms.

4_x_ + 5 = 13_x_ + 4 – x – 9

4_x_ + 5 = 12_x_ – 5

–8_x_ = –10

x = 5/4

First we solve for x.

Subtracting 3 from both sides gives us –3_x_ < 17.

Dividing by –3 gives us x > –17/3.

–6 is less than –17/3.

The problem tells us that increasing x by twenty percent gives us the same thing that we would get if we decreased the product of four and x by five. We need to find expressions for these two situations, and then we can set them equal and solve for x.

Let's find an expression for increasing x by twenty percent. We could represent this as x + 20%x = x + 0.2x = 1.2x = 6x/5.

Let's find an expression for decreasing the product of four and x by five. First, we must find the product of four and x, which can be written as 4x. Then we must decrease this by five, so we must subtract five from 4x, which could be written as 4x - 5.

Now we must set the two expressions equal to one another.

6x/5 = 4x - 5

Subtract 6x/5 from both sides. We can rewrite 4x as 20x/5 so that it has a common denominator with 6x/5.

0 = 20x/5 - 6x/5 - 5 = 14x/5 - 5

0 = 14x/5 - 5

Now we can add five to both sides.

5 = 14x/5

Now we can multiply both sides by 5/14, which is the reciprocal of 14/5.

5(5/14) = (14x/5)(5/14) = x

25/14 = x

The answer is 25/14.

This question gives two equalities and one inequality. The inequality (sv ≠ 0) simply says that neither s nor v is 0. The two equalities tell us that 4_xs and ks are both equal to v, which means that 4_xs_ and ks must be equal to each other--that is, 4_xs_ = ks. Dividing both sides by s gives 4_x_ = k, which is our solution.

We are looking for the value of t that gives $675 as the result when plugged in V (t ). While there are many ways to do this, one of the fastest is to plug in the answer choices as values of t .

When we plug t = 1 into _V (_t ), we get V (1) = 1200(0.75)1 = 1000(0.75) = $900, which is incorrect.

When we plug t = 2 into V (t ), we get V (2) = 1200(0.75)2 = $675, so this is our solution.

The value of the tractor will be $675 after 2 years.

Finally, we can see that if t = 3, 4, or 5, the resulting values of the V (t ) are all incorrect.

Since we are told that 2x2(5-x)(3x+2) = 0, in order to find x, we must let each of the factors of our equation equal zero. The equation is already factored for us, which means that our factors are 2x2, (5-x), and (3x+2). We must let each of these equal zero separately, and these will give us the possible values of x that satisfy the equation.

Let's look at the factor 2x2 and set it equal to zero.

2x2 = 0

x = 0

Then, let's look at the factor 5-x.

5-x = 0

Add x to both sides

5 = x

x = 5

Finally, we set the last factor equal to zero.

(3x+2) = 0

Subtract two from both sides

3x = -2

Divide both sides by three.

x = -2/3

This means that the possible values of x are 0, 5, or -2/3. The question asks us to find the sum of these values.

0 + 5 + -2/3

5 + -2/3

Remember to find a common denominator of 3.

15/3 + -2/3 = 13/3.

The answer is 13/3.