All SAT Math Resources
Example Questions
Example Question #271 : Equations / Inequalities
How many solutions are there to the following system of equations?
There are no solutions.
There are 2 solutions.
There are 3 solutions.
There is 1 single solution.
There are an infinite number of solutions.
There are an infinite number of solutions.
If we use elimination to solve this system of equations, we can add the two equations together. This results in 0=0.
When elimination results in 0=0, that means that the two equations represent the same line. Therefore, there are an infinite number of solutions.
Example Question #272 : Algebra
Solve the system of equations:
None of the given answers.
We can solve this system of equations by elimination since the 2 given y-values have the same coefficient. Let's subtract the second equation from the first.
This gives us or .
Substitute this x-value into either equation and solve for y. Let's use the first equation like so:
The solution is .
Example Question #61 : Systems Of Equations
Solve the system of equations given below.
We can solve this problem by elimination.
Subtract the second equation from the first to eliminate the x values, like so:
This yields
Now, substitute into either equation and solve for x. In either equation, we find that .
Therefore, the solution to this system of equations is .
Example Question #61 : Systems Of Equations
At the cafe, Jalynn buys five donuts and a cup of coffee for . Giovanni buys five donuts and two cups of coffee for . How much does each cup of coffee cost?
To solve this problem, we can set up a system of equations.
Let represent the cost of donuts, and let represent the cost of a cup of coffee.
We can write Giovanni and Jalynn's purchases as equations like this:
To solve this system of equations, we can use elimination by subtracting the second equation from the first, like so:
Our terms drop out, leaving us with
Therefore, each cup of coffee costs .
Example Question #62 : Systems Of Equations
Solve the system of equations:
Because the coefficients of the values are the same, we can solve this problem by elimination. Subtract the second equation from the first:
This knocks out our value and gives us , or .
Then, substitute this value into either of the equations to solve for . Let's use the second equation:
Therefore, the solution is
Example Question #61 : Systems Of Equations
Solve the system of equations:
Since none of our coefficients are the same, we cannot solve by elimination. Therefore, we'll use substitution. Let's solve for one term in one equation.
Now, we can substitute this expression into the second equation.
Now, we can substitute this value back into either of the original equations to find our solution. Let's use the simplest equation:
With these values, we see that our solution is the point .
Example Question #63 : Systems Of Equations
Evaluate .
The system has no solution.
Set up the system:
Multiply both sides of the first equation by 5, and both sides of the second by :
Add both sides of the equations, then solve for in the resulting equation:
Example Question #64 : Systems Of Equations
Evaluate .
The system has no solution.
In the equation , substitute for :
Simplify, then solve for by isolating the variable:
Example Question #281 : Algebra
Evaluate .
The system has no solution.
Multiply both sides of each equation by the least common denominators of their coefficients in order to make those coefficients into whole numbers:
, so
, so
This sets up the system
Multiply the latter equation by on both sides, then add to the former equation to eliminate the terms:
Solve for :
Example Question #281 : Algebra
Evaluate .
The system has no solution.
In the equation , substitute for :
Simplify the left expression:
Now solve for :
Substitute this value in the following equation and evaluate for :