Systems of Equations - Algebra

To factor a quadratic equation in the form

, where , find two integers that have a sum of and a product of .

For this equation, that would be 9 and 5.

Therefore, the solutions to this equation are and .

1. After adding like terms and setting the equation equal to zero, the immediate next step in solving any quadratic is to simplify. If the coefficients of all three terms have a common factor, pull it out. So go ahead and divide both sides (and therefore ALL terms on BOTH sides) by 4.

Since zero divided by four is still zero, only the left side of the equation changes.

1. Either factor by grouping or use the square trick.

Grouping:

1 + 1 = 2

(The "1" was pulled out only to make the next factoring step clear.)

x + 1 = 0, x = –1

OR

Perfect Square:

x = –1

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.

The question is asking the student to solve the linear set of equations ultimately by isolating and .

There are a few ways a student could choose to answer this.

One may see immediately and realize to eliminate the result from the second equation one would subtract as follows.

--

Other students may choose to subtract the first equation twice from the second equation then subsequently solve for y.

To solve this system of equations, multiply each equation by an integer that would allow us to cancel one variable. We then substitute that value back into either original equation to find the other variable. In this case we will cancel the x variable by multiplying each equation by 3 and 4 respectively.

Multiply equation 1 by 3 and equation 2 by 4:

Add the new equations:

The result is

Solve for y:

Now substitute this value back into equation 1:

Because the two rational expressions have the same denominator, we can simply add straight across the top. The denominator stays the same.

Therefore the answer is .

Multiply both sides by :

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is . These integers are :

Set each factor equal to 0 and solve separately:

or

Notice that the end value is a negative. Any negative or positive value that is inside an absolute value sign must result to a positive value.

If we split the equation to its positive and negative solutions, we have:

Solve the first equation.

The answer to is:

Solve the second equation.

The answer to is:

If we substitute these two solutions back to the original equation, the results are positive answers and can never be equal to negative one.

The answer is no solution.

The key to solving this question is noticing that we can factor out a 2:

2_x_ + 6_y_ = 44 is the same as 2(x + 3_y_) = 44.

Therefore, x + 3_y_ = 22.

In this case, x + 3_y_ + 33 is the same as 22 + 33, or 55.

With rational equations we must first note the domain, which is all real numbers except and . That is, these are the values of that will cause the equation to be undefined. Since the least common denominator of , , and is , we can mulitply each term by the LCD to cancel out the denominators and reduce the equation to . Combining like terms, we end up with . Dividing both sides of the equation by the constant, we obtain an answer of . However, this solution is NOT in the domain. Thus, there is NO SOLUTION because is an extraneous answer.

Combine like terms on each side of the equation:

Next, subtract from both sides.

Then subtract from both sides.

This is nonsensical; therefore, there is no solution to the equation.

Use the substitution method to solve for the solution set.

Solve equation 2 for y:

Substitute into equation 1:

If equation 1 was solved for a variable and then substituted into the second equation a similar result would be found. This is because these two equations have No solution. Change both equations into slope-intercept form and graph to visualize. These lines are parallel; they cannot intersect.

*Any method of finding the solution to this system of equations will result in a no solution answer.

When finding how many solutions an equation has you need to look at the constants and coefficients.

The coefficients are the numbers alongside the variables.

The constants are the numbers alone with no variables.

If the coefficients are the same on both sides then the sides will not equal, therefore no solutions will occur.

Use distributive property on the right side first.

No solutions

First factorize the numerator.

Rewrite the equation.

The terms can be eliminated.

Subtract one on both sides.

However, let's substitute this answer back to the original equation to check whether if we will get as an answer.

Simplify the left side.

The left side does not satisfy the equation because the fraction cannot be divided by zero.

Therefore, is not valid.

The answer is:

With rational equations we must first note the domain, which is all real numbers except . (If , then the term will be undefined.) Next, the least common denominator is , so we multiply every term by the LCD in order to cancel out the denominators. The resulting equation is . Subtract on both sides of the equation to collect all variables on one side: . Lastly, divide by the constant to isolate the variable, and the answer is . Be sure to double check that the solution is in the domain of our equation, which it is.

With rational equations we must first note the domain, which is all real numbers except . (Recall, the denominator cannot equal zero. Thus, to find the domain set each denominator equal to zero and solve for what the variable cannot be.)

The least common denominator or and is . Multiply every term by the LCD to cancel out the denominators. The equation reduces to . We can FOIL to expand the equation to . Combine like terms and solve: . Factor the quadratic and set each factor equal to zero to obtain the solution, which is or . These answers are valid because they are in the domain.

Before you can add the two numerators from the fractions, the fractions must have a common denominator. The common denominator will be the Lowest Common Denominator (LCD). When you are dealing with variables, you get this by multiplying the two denominators together:

To follow through with this LCD, you must multiply each fraction by the other fraction's denominator so that they end up with common denominators. The first fraction needs to be multiplied by and the second fraction needs to be multiplied by . Note that both of these are equivalent to ONE, so the value of the fraction does not change:

Now that the two fractions have common denominators, you can add the two numerators:

Before you can subtract the two numerators from the fractions, the fractions must have a common denominator. The common denominator will be the Lowest Common Denominator (LCD). When you are dealing with variables, you get this by multiplying the two denominators together:

To follow through with this LCD, you must multiply each fraction by the other fraction's denominator so that they end up with common denominators. The first fraction needs to be multiplied by and the second fraction needs to be multiplied by . Note that both of these are equivalent to ONE, so the value of the fraction does not change:

Now that the two fractions have common denominators, you can subtract the two numerators. You can also distribute the denominator so you can simplify later:

Finally you can divide out each term in the numerator and denominator by 2 to fully simplify the answer:

Since the two fractions already have a common denominator, you can add the fractions by adding up the two numerators and keeping the common denominator:

Next you will algebraically solve for by isolating it on one side of the equation. The first step is to multiply each side by :

Cancel out the on the left and distribute out on the right. Then solve for by subtracting to the left and subtracting 10 to the right. Finally divide each side by negative 2:

There are two ways to solve this problem using LCD (Least Common Denominator). The first method requires that you convert all denominators to the LCD by multiplying appropriately, and then follow the operations the equation requests. The second method allows you to cancel out terms using the LCD by mutiplying each term by the LCD. This is the method used for this problem and sometimes the simpler method since it tends to eliminate some of the fractions.

Find the LCD:

The LCD will be based of the denominator of the first fraction (with 8 in the numerator). The middle term, based on this, is missing an x while the third term is missing . Our LCD will be since it has all the parts of each denominator. Multiply each term by this LCD.

Cancel terms that are present in our denominators and the LCD (same terms) we are multiplying by (red numbers will be canceled out):

Rewrite equation:

Distribute the right side of the equation:

Move x's to one side:

completely isolate x: