How to find a solution set - Algebra

Card 0 of 171

Question

Solve for :

$x = \sqrt{2x-1} +8$

Answer

Isolate the radical, square both sides, and solve the resulting quadratic equation.

$x = \sqrt{2x-1} +8$

$x - 8 = \sqrt{2x-1} +8 -8$

$x - 8 = \sqrt{2x-1}$

$\left (x -8 \right )^{2} =\left ( \sqrt{2x-1} \right )^{2}$

$x^{2} - 2 \cdot x \cdot 8 + 8^{2} = 2x - 1$

$x^{2} - 16 x + 64 = 2x - 1$

$x^{2} - 16 x + 64 - 2x + 1 = 2x - 1 - 2x + 1$

$x^{2} - 18 x + 65 = 0$

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 .

$x = \sqrt{2x-1} +8$

$5 = \sqrt{2 \cdot 5-1} +8$

$5 = \sqrt{9} +8$

This is a false statement, so 5 is a false "solution".

$x = \sqrt{2x-1} +8$

$13 = \sqrt{2 \cdot 13-1} +8$

$13= \sqrt{25} +8$

This is a true statement, so 13 is the only solution of the equation.

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Question

Solve this system of equations.

Answer

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.

$z=-\frac{5}{3}$

We can use this z value to find y

$y=-3\left ( -\frac{5}{3}\right )-3$

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

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Question

Give all real solutions of the following equation:

$x^{4} +5x^{2} - 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} +5x^{2} - 36 = 0$

$u^{2} + 5u - 36 = 0$

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

$(x^{2}+9)(x^{2}-4) = 0$

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

$(x^{2}+9)(x+2)(x-2) = 0$

Set each factor to zero and solve:

$x^{2}+9= 0$

$x^{2}= -9$

Since no real number squared is equal to a negative number, no real solution presents itself here.

$x-2= 0 \Rightarrow x = 2$

$x+2 = 0 \Rightarrow x = - 2$

The solution set is $\left { -2, 2\right }$ .

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Question

Solve for :

Answer

Subtract 12 from both sides:

–3x = –15

Divide both sides by –3:

x = 5

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Question

Solve for :

Answer

Distribute the x through the parentheses:

x2 –2x = x2 – 8

Subtract x2 from both sides:

–2x = –8

Divide both sides by –2:

x = 4

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Question

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?

Answer

To make this much easier, translate the word problem into a system of three equations.

$N=\frac{1}{4}S-1$

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.

$N=\frac{1}{4}(8)-1=1$

Newton is one year old. So the answer is:

Sora, 8 years

Newton, 1 year

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Question

Solve for : $x^{2}-144=0$

Answer

To solve this problem we can first add to each side of the equation yielding $x^{2}=144$

Then we take the square root of both sides to get $\sqrt{x^{2}}=\sqrt{144}$

Then we calculate the square root of which is $x=\pm12$ .

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Question

Determine where the graphs of the following equations will intersect.

Answer

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 .

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Question

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?

Answer

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.

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

Finally, count the number of multiples found in both sets. Both sets include the following numbers:

$\text{12, 24, 36, and 48}$

The correct answer is 4 numbers.

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Question

Give all real solutions of the following equation:

$x^{4} - 13x^{2} + 36 = 0$

Answer

By substituting $u = x^{2}$ - and, subsequently, $u^{2} =\left ( x^{2} \right )^{2} = x^{4}$ this can be rewritten as a quadratic equation, and solved as such:

$x^{4} - 13x^{2} + 36 = 0$

$u^{2} - 13u + 36 = 0$

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:

$(x^{2}-9)(x^{2}-4) = 0$

These factors can themselves be factored as the difference of squares:

Set each factor to zero and solve:

$x-3 = 0 \Rightarrow x = 3$

$x-2= 0 \Rightarrow x = 2$

$x+2 = 0 \Rightarrow x = - 2$

$x+3 = 0 \Rightarrow x = - 3$

The solution set is $\left { -3, -2, 2, 3\right }$ .

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Question

Solve for :

$\frac{6}{x-2} + 1 = x$

Answer

Subtract 1 from both sides, then multiply all sides by :

$\frac{6}{x-2} + 1 = x$

$\frac{6}{x-2} + 1 - 1 = x- 1$

$\frac{6}{x-2} = x- 1$

$\frac{6}{x-2} \cdot (x-2) = \left (x- 1 \right ) \cdot (x-2)$

$6 = x^{2} -3x + 2$

$6 - 6 = x^{2} -3x + 2 - 6$

$0 = x^{2} -3x -4$

A quadratic equation is yielded. We can factor the expression, then set each individual factor to 0.

$x^{2} -3x -4 =0$

$x-4 = 0 \Rightarrow x = 4$

$x+1 = 0 \Rightarrow x = -1$

Both of these solutions can be confirmed by substitution.

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Question

Solve for :

$\sqrt[3]{2n-8} = -4$

Answer

Cube both sides of the equation to form a linear equation, then solve:

$\sqrt[3]{2n-8} = -4$

$\dpi{150} \left (\sqrt[3]{2n-8} \right ) ^{3}= \left (-4 \right )^{3}$

$2n \div 2 = -56\div 2$

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Question

Solve for :

$\dpi{130} \sqrt[3]{100-x} - 5 = 0$

Answer

Add 5 to both sides to isolate the cube root:

$\dpi{130} \sqrt[3]{100-x} - 5+ 5 = 0+ 5$

$\dpi{130} \sqrt[3]{100-x} = 5$

Cube both sides:

$\dpi{130} \left (\sqrt[3]{100-x} \right) ^{3}= 5 ^{3}$

To isolate , move it to the right side of the equation. We choose the right side instead of the left side so as to make positive:

Subtract 125 from both sides to isolate :

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Question

Solve for :

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

Answer

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.

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Question

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

Answer

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.

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Question

Solve the equation for .

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

Answer

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

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

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Question

Find the solution, in the form , to the following system of equations:

Answer

Multiply (1) by 3, multiply (2) by 4:

$3\cdot (4x+3y=9)=12x+9y=27$

$4\cdot(-3x-5y=-26)=-12x-20y=-104$

Add the two resulting solutions:

Substitute into (1) and solve for :

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Question

Answer

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:

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Question

Solve the following by substitution:

Answer

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.

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Question

Solve for :

$x = \sqrt{2x-1} +8$

Answer

Isolate the radical, square both sides, and solve the resulting quadratic equation.

$x = \sqrt{2x-1} +8$

$x - 8 = \sqrt{2x-1} +8 -8$

$x - 8 = \sqrt{2x-1}$

$\left (x -8 \right )^{2} =\left ( \sqrt{2x-1} \right )^{2}$

$x^{2} - 2 \cdot x \cdot 8 + 8^{2} = 2x - 1$

$x^{2} - 16 x + 64 = 2x - 1$

$x^{2} - 16 x + 64 - 2x + 1 = 2x - 1 - 2x + 1$

$x^{2} - 18 x + 65 = 0$

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 .

$x = \sqrt{2x-1} +8$

$5 = \sqrt{2 \cdot 5-1} +8$

$5 = \sqrt{9} +8$

This is a false statement, so 5 is a false "solution".

$x = \sqrt{2x-1} +8$

$13 = \sqrt{2 \cdot 13-1} +8$

$13= \sqrt{25} +8$

This is a true statement, so 13 is the only solution of the equation.

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