Equations / Solution Sets - Algebra 1

First pull out 3u from both terms.

3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is _a_3 – _b_3 = (a – b)(_a_2 + ab + b_2). In our problem, a = u and b = 2_v:

3_u_4 – 24_uv_3 = 3_u_(u_3 – 8_v_3) = 3_u\[u_3 – (2_v)3\]

= 3_u_(u – 2_v_)(u_2 + 2_uv + 4_v_2)

Add 8 to both sides to set the equation equal to 0:

To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

Then factor by grouping:

Set each factor equal to 0 and solve:

and

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

The given expression is a special binomial, known as the "difference of squares". A difference of squares binomial has the given factorization: . Thus, we can rewrite as and it follows that

The product of is .

For the equation ,

must equal and must equal .

Thus and must be and , making the answer .

This is a quadratic equation. We can solve for either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format .

To yield the first value in our original equation (), and .

To yield the final term in our original equation (), we can set and .

Now that the equation has been factored, we can evaluate . We set each factored term equal to zero and solve.

First, find the factors of 90 and 315. The greatest common factor is the largest factor shared by both of the numbers: 45.

Then, find the least common multiple of 5 and 15. This will be the smallest number that can be divided by both 5 and 15: 15.

Finally, the greatest common factor (45) divided by the least common multiple (15) = 45 / 15 = 3.

First factor the numerator. We need two numbers with a sum of 3 and a product of 2. The numbers 1 and 2 satisfy these conditions:

Now, look to see if there are any common factors that will cancel:

The in the numerator and denominator cancel, leaving .

The general form for a factored expression of order 2 is

, which, when FOILED, gives .

Comparing this generic expression to the one given in the probem, we can see that the term should equal , and the term should equal 2.

The values of and that satisfy the two equations are and ,

so your factored expression is

The first step in solving the equation is to add to both sides so that we can set the new equation equal to so that we can factor it.

We now need to factor it. We need to find two numbers, and , such that and .

Those two numbers are

Our new factored expression becomes

Now we can easily identify that the values that satisfy this equation are .

In order to set this up for completing the square, we need to move the 135 to the other side:

Now the equation is in the form:

To complete the square we need to add to both sides the folowing value:

So we need to add 9 to both sides of the equation:

Now we can factor the left side and simplify the left side:

Now we need to take the square root of both sides:

**NOTE: Don't foget to add the plus or minus symbol. We add this becase there are two values we can square to get 144:

and

**End note

Now we can split into two equations and solve for x:

and

So our solution is:

There is a common variable in each term. Pull this out as a common factor.

Factor .

The common factors that will achieve the middle term and will have a product of 36 is four and nine. Write the binomials.

The cubics here cannot be simplified any further.

The answer is:

Subtract 12 from both sides:

–3x = –15

Divide both sides by –3:

x = 5

Distribute the x through the parentheses:

x2 –2x = x2 – 8

Subtract x2 from both sides:

–2x = –8

Divide both sides by –2:

x = 4

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

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

Then, add that to the third equation so that the y and z terms are eliminated. You will get .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

We can use this z value to find y

So the solution set is x = 1, y = 2, and z = –5/3.

To solve this problem we can first add to each side of the equation yielding

Then we take the square root of both sides to get

Then we calculate the square root of which is .

We can solve the system of equations using the substitution method.

Solve for in the second equation.

Substitute this value of into the first equation.

Now we can solve for .

Solve for using the first equation with this new value of .

The solution is the ordered pair .

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.