Algebra - How to find a solution set

Solve the following by substitution:

No solution

Explanation

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.

No solution.

None of the other answers.

Explanation

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:

Solve the equation for .

$\left | 5x-4\right |=\left | 2x+7\right |$

$x=\frac{11}{3};or;x=-\frac{3}{7}$

$x=\frac{3}{7}:: or: :x=-\frac{11}{3}$

$x=\frac{5}{8};or;x=4$

$x=\frac{11}{7};or;x=-\frac{11}{7}$

Explanation

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.

$x=\frac{11}{3}$ $x=-\frac{3}{7}$

If the area of a rectangle is 100 square feet and the width is 20 feet, then what is the perimeter?

50 feet

80 feet

30 feet

20 feet

Explanation

The area of a rectangle is , where A is the area, L is the length, and W is the width. The perimeter is given by . We know that and . We can solve for L using $L=\frac{A}{W}=\frac{100}{20}=5.$ The perimeter is then feet.

Determine where the graphs of the following equations will intersect.

Explanation

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 .

$y=\frac{14}{-7}=-2$

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

$x=\frac{3}{3}=1$

The solution is the ordered pair .

Solve for :

$\sqrt{x} + \sqrt{x -5} = 5$

$x = 4 \textrm{ or }x = 9$

$x = -4 \textrm{ or }x = 9$

The equation has no solution.

Explanation

Move one radical to the other side, then square, thereby yielding an equation with only one radical.

$\sqrt{x} + \sqrt{x -5} = 5$

$\sqrt{x} + \sqrt{x -5} -\sqrt{x} = 5-\sqrt{x}$

$\sqrt{x -5} = 5-\sqrt{x}$

$\left (\sqrt{x -5} \right ) ^{2} =\left ( 5-\sqrt{x }\right ) ^{2}$

$x -5= 5 ^{2} - 2 \cdot 5 \cdot \sqrt{x } + \left ( \sqrt{x }\right ) ^{2}$

$x -5= 25 - 10 \sqrt{x } + x$

Isolate the radical on one side, then square.

$x -5-x -25= 25 - 10 \sqrt{x } + x -x -25$

$-30 = - 10 \sqrt{x }$

$-30 \div (-10) = - 10 \sqrt{x } \div (-10)$

$3= \sqrt{x }$

$3^{2} =\left ( \sqrt{x } \right )^{2}$

Substitution confirms this to be the only solution.

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?

Explanation

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:

$\text{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, and 48}$

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.

$\text{6, 12, 18, 24, 30, 36, 42, and 48}$