Linear / Rational / Variable Equations - SAT Math
Card 0 of 960
Find the solution to the following equation if x = 3:
y = (4x2 - 2)/(9 - x2)
Find the solution to the following equation if x = 3:
y = (4x2 - 2)/(9 - x2)
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.
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.
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I. x = 0
II. x = –1
III. x = 1
I. x = 0
II. x = –1
III. x = 1
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A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.
A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.
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Solve:
Solve:
First, distribute, making sure to watch for negatives.
Combine like terms.
Subtract 7x from both sides.
Add 18 on both sides and be careful adding integers.
First, distribute, making sure to watch for negatives.
Combine like terms.
Subtract 7x from both sides.
Add 18 on both sides and be careful adding integers.
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Solve:
Solve:
First, distribute the 
to the terms inside the parentheses.
Add 6x to both sides.
This is false for any value of 
. Thus, there is no solution.
First, distribute the
Add 6x to both sides.
This is false for any value of
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Solve 
.
Solve
By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.
By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.
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, 
In the above graphic, approximately determine the x values where the graph is neither increasing or decreasing.
,
In the above graphic, approximately determine the x values where the graph is neither increasing or decreasing.
We need to find where the graph's slope is approximately zero. There is a straight line between the x values of 
, and 
. The other x values have a slope. So our final answer is 
.
We need to find where the graph's slope is approximately zero. There is a straight line between the x values of
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In the equation below, 
, 
, and 
are non-zero numbers. What is the value of 
in terms of 
and 
?
In the equation below,
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Solve for x:
Solve for x:
The first step is to cancel out the denominator by multiplying both sides by 7:
Subtract 3 from both sides to get 
by itself:
The first step is to cancel out the denominator by multiplying both sides by 7:
Subtract 3 from both sides to get
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Solve for 
and 
using elimination:
Solve for
When using elimination, you need two factors to cancel out when the two equations are added together. We can get the 
in the first equation to cancel out with the 
in the second equation by multiplying everything in the second equation by 
:
Now our two equations look like this:
The 
can cancel with the 
, giving us:
These equations, when summed, give us:
Once we know the value for 
, we can just plug it into one of our original equations to solve for the value of 
:
When using elimination, you need two factors to cancel out when the two equations are added together. We can get the
Now our two equations look like this:
The
These equations, when summed, give us:
Once we know the value for
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Give the solution set of the rational equation 
Give the solution set of the rational equation
Multiply both sides of the equation by the denominator 
:
Rewrite both expression using the binomial square pattern:
This can be rewritten as a linear equation by subtracting 
from both sides:
Solve as a linear equation:
Multiply both sides of the equation by the denominator
Rewrite both expression using the binomial square pattern:
This can be rewritten as a linear equation by subtracting
Solve as a linear equation:
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Solve:
Solve:
Multiply by 
on each side
Subtract 
on each side
Multiply by 
on each side
Multiply by
Subtract
Multiply by
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If 6_x_ = 42 and xk = 2, what is the value of k?
If 6_x_ = 42 and xk = 2, what is the value of k?
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.
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.
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If 4_x_ + 5 = 13_x_ + 4 – x – 9, then x = ?
If 4_x_ + 5 = 13_x_ + 4 – x – 9, then x = ?
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
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
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If 3 – 3_x_ < 20, which of the following could not be a value of x?
If 3 – 3_x_ < 20, which of the following could not be a value of x?
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.
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.
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If 
, then, in terms of 
, 
If
You can solve this problem by plugging in random values or by simply solving for k. To solve for k, put the s values on one side and the k values on the other side of the equation. First, subtract 4s from both sides. This gives 4s – 6k = –2k. Next, add 6k to both sides. This leaves you with 4s = 4k, which simplifies to s=k. The answer is therefore s.
You can solve this problem by plugging in random values or by simply solving for k. To solve for k, put the s values on one side and the k values on the other side of the equation. First, subtract 4s from both sides. This gives 4s – 6k = –2k. Next, add 6k to both sides. This leaves you with 4s = 4k, which simplifies to s=k. The answer is therefore s.
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What is the value of (5 + x)(10 – y) when x = 3 and y = –3?
What is the value of (5 + x)(10 – y) when x = 3 and y = –3?
This is a simple plug-in and PEMDAS problem. First, plug in x = 3 and y = –3 into the x and y. You should follow the orders of operation and compute what is within the parentheses first and then find the product. This gives 8 * 13 = 104. The answer is 104.
This is a simple plug-in and PEMDAS problem. First, plug in x = 3 and y = –3 into the x and y. You should follow the orders of operation and compute what is within the parentheses first and then find the product. This gives 8 * 13 = 104. The answer is 104.
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