All Algebra 1 Resources
Example Questions
Example Question #2 : Expressions
We have three cats, Chai, Sora, and Newton. Chai is 3 years old. Sora two years older than twice Chai's age. Newton is one year younger than one-fourth of Sora's age. How old are Sora and Newton?
Sora: 8 years
Newton: 1 year
Sora: 1 year
Newton: 8 years
Sora: 3 years
Newton: not born yet
Sora: 5 years
Newton: 6 years
Sora: 4 years
Newton: one half year
Sora: 8 years
Newton: 1 year
To make this much easier, translate the word problem into a system of three equations.
We have C for Chai, S for Sora, and N for Newton. To find Sora's age, plug in into .
Sora is 8 years old. Use this to find Newton's age.
Newton is one year old. So the answer is:
Sora, 8 years
Newton, 1 year
Example Question #11 : How To Find A Solution Set
Give all real solutions of the following equation:
The equation has no real solutions.
By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:
We are looking to factor the quadratic expression as , replacing the two question marks with integers with product 36 and sum ; these integers are .
Substitute back:
These factors can themselves be factored as the difference of squares:
Set each factor to zero and solve:
The solution set is .
Example Question #151 : Rational Expressions
Solve for :
There is no solution.
Subtract 1 from both sides, then multiply all sides by :
A quadratic equation is yielded. We can factor the expression, then set each individual factor to 0.
Both of these solutions can be confirmed by substitution.
Example Question #12 : How To Find A Solution Set
Solve for :
Isolate the radical, square both sides, and solve the resulting quadratic equation.
Factor the expression at left by finding two integers whose product is 65 and whose sum is ; they are .
Set each linear binomial to 0 and solve for to find the possible solutions.
or
Substitute each for .
This is a false statement, so 5 is a false "solution".
This is a true statement, so 13 is the only solution of the equation.
Example Question #261 : Equations / Inequalities
Solve the equation for .
The two sides of the equation will be equal if the quantities inside the absolute value signs are equal or equal with opposite signs.
From this, you get two equations.
or
Example Question #266 : Equations / Inequalities
Find the solution, in the form , to the following system of equations:
Multiply (1) by 3, multiply (2) by 4:
Add the two resulting solutions:
Substitute into (1) and solve for :
Example Question #267 : Equations / Inequalities
None of the other answers.
No solution.
To find the solution isolate the variable on one side of the equation.
To check to see if this is the correct solution, plug the value of x back into the equation and solve:
Example Question #271 : Equations / Inequalities
Solve the following by substitution:
No solution
To solve this equation you will plug the second equation straight into the first one by substituting what is written there for .
You will then get:
From here you need to simplify by combining like terms:
Bring the over by addition:
Then divide both sides by to get:
You will then take this value of and plug it into either equation.
Example Question #271 : Equations / Inequalities
Set A is composed of all multiples of 4 that are that are less than the square of 7. Set B includes all multiples of 6 that are greater than 0. How many numbers are found in both set A and set B?
Start by making a list of the multiples of 4 that are smaller than the square of 7. When 7 is squared, it equals 49; thus, we can compose the following list:
Next, make a list of all the multiples of 6 that are greater than 0. Since we are looking for shared multiples, stop after 48 because numbers greater than 48 will not be included in set A. The biggest multiple of 4 smaller that is less than 49 is 48; therefore, do not calculate multiples of 6 greater than 48.
Finally, count the number of multiples found in both sets. Both sets include the following numbers:
The correct answer is 4 numbers.
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