✅ Formula to find x ⭐️⭐️⭐️⭐️⭐

How to Find X in an Algebra Question

Algebra is a type of mathematics that introduces the concept of variables representing numbers. “X” is one such variable used in algebraic equations. You can find “x” or solve the equation for “x” by isolating the “x” on one side of the algebraic equation. To solve for “x”, you need to understand the basic rules of algebraic operations.

Isolate “x” on one side of the algebraic equation by subtracting the sum that appears on the same side of the equation as the “x.” For example, in the equation “x + 5 = 12”, rewrite the equation as “x = 12 – 5” and solve for “x.” The solution is “x = 7.”

Isolate “x” on one side of the algebraic equation by adding the negative number that appears on the same side of the equation as the “x.” For example, in the equation “x – 5 = 12”, rewrite the equation as “x = 12 + 5” and solve for “x.” The solution is “x = 17.”

Isolate “x” on one side of the algebraic equation by dividing the number that appears on the same side of the equation as part of “x.” For example, in the equation “12x = 24”, rewrite the equation as “x = 24 / 12” and solve for “x.” The solution is “x = 2.”

Isolate “x” on one side of the algebraic equation by multiplying the number that appears on the same side of the equation as part of an “x” fractional component. For example, in the equation “x / 2 = 3, rewrite the equation as “x = 2 x 3” and solve for “x.” The solution is “x = 6.”

How to Solve for X

Using a Basic Linear Equation

Write down the problem. Here it is:

• 2²(x+3) + 9 – 5 = 32

Resolve the exponent. Remember the order of operations: PEMDAS, which stands for Parentheses, Exponents, Multiplication/Division, and Addition/Subtraction. You can’t resolve the parentheses first because x is in the parentheses, so you should start with the exponent, 2². 2² = 4

• 4(x+3) + 9 – 5 = 32

Do the multiplication.Just distribute the 4 into (x +3). Here’s how:

• 4x + 12 + 9 – 5 = 32

Do the addition and subtraction. Just add or subtract the remaining numbers. Here’s how:

• 4x+21-5 = 32
• 4x+16 = 32
• 4x + 16 – 16 = 32 – 16
• 4x = 16

Isolate the variable To do this, just divide both sides of the equation by 4 to find x. 4x/4 = x and 16/4 = 4, so x = 4.

• 4x/4 = 16/4
• x = 4

Check your work. Just plug x = 4 back into the original equation to make sure that it checks out. Here’s how:

• 2²(x+3)+ 9 – 5 = 32
• 2²(4+3)+ 9 – 5 = 32
• 2²(7) + 9 – 5 = 32
• 4(7) + 9 – 5 = 32
• 28 + 9 – 5 = 32
• 37 – 5 = 32
• 32 = 32

With Exponents

Write down the problem. Let’s say you’re working with this problem where the x term includes an exponent:

• 2x² + 12 = 44

Isolate the term with the exponent. The first thing you should do is combine like terms so that all of the constant terms are on the right side of the equation while the term with the exponent is on the left side. Just subtract 12 from both sides. Here’s how:

• 2x²+12-12 = 44-12
• 2x² = 32

Isolate the variable with the exponent by dividing both sides by the coefficient of the x term. In this case, 2 is the x coefficient, so divide both sides of the equation by 2 to get rid of it. Here’s how:

• (2x²)/2 = 32/2
• x² = 16

Take the square root of each side of the equation. Taking the square root of x² will cancel it out. So, take the square root of both sides. You’ll get x left over on one side and plus or minus the square root of 16, 4, on the other side. Therefore, x = ±4.

Check your work. Just plug x = 4 and x = -4 back in to the original equation to make sure it checks out. For instance, when you’re checking x=4:

• 2x² + 12 = 44
• 2 x (4)² + 12 = 44
• 2 x 16 + 12 = 44
• 32 + 12 = 44
• 44 = 44

Using Fractions

Write down the problem. Let’s say you’re working with the following problem:

• (x + 3)/6 = 2/3

Cross multiply. To cross multiply, simply multiply the denominator of each fraction by the numerator of the other fraction. You’ll be essentially multiplying in two diagonal lines. So, multiply the first denominator, 6, by the second numerator, 2, to get 12 on the right side of the equation. Multiply the second denominator, 3, by the first numerator, x + 3, to get 3 x + 9 on the left side of the equation. Here’s how it will look:

• (x + 3)/6 = 2/3
• 6 x 2 = 12
• (x + 3) x 3 = 3x + 9
• 3x + 9 = 12

Combine like terms. Combine the constant terms in the equation to subtract 9 from both sides of the equation. Here’s what you do:

• 3x + 9 – 9 = 12 – 9
• 3x = 3

Isolate x by dividing each term by the x coefficient. Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you’re left with x = 1.

Check your work. To check your work, just plug x back in to the original equation to make sure that it works. Here’s what you do:

• (x + 3)/6 = 2/3
• (1 + 3)/6 = 2/3
• 4/6 = 2/3
• 2/3 = 2/3

Using Radical Signs

Write down the problem. Let’s say you’re solving for x in the following problem:

• √(2x+9) – 5 = 0

Isolate the square root. You have to move the part of the equation with the square root sign to one side of the equation before you can proceed. So, you’ll have to add 5 to both sides of the equation. Here’s how:

• √(2x+9) – 5 + 5 = 0 + 5
• √(2x+9) = 5

Square both sides. Just as you would divide both sides of an equation by a coefficient that is being multiplied by x, you would square both sides of an equation if x appears under the square root, or the radical sign. This will remove the radical sign from the equation. Here’s how you do it:

• (√(2x+9))² = 5²
• 2x + 9 = 25

Combine like terms. Combine like terms by subtracting both sides by 9 so that all of the constant terms are on the right side of the equation while x remains on the left side. Here’s what you do:

• 2x + 9 – 9 = 25 – 9
• 2x = 16

Isolate the variable. The last thing you have to do to solve for x is to isolate the variable by dividing both sides of the equation by 2, the coefficient of the x term. 2x/2 = x and 16/2 = 8, so you’re left with x = 8.

Check your work. Plug 8 back in to the equation for x to see if you get the right answer:

• √(2x+9) – 5 = 0
• √(2(8)+9) – 5 = 0
• √(16+9) – 5 = 0
• √(25) – 5 = 0
• 5 – 5 = 0

Using Absolute Value

Write down the problem. Let’s say you’re trying to solve for x in the following problem:

• |4x +2| – 6 = 8

Isolate the absolute value. The first thing you have to do is to combine like terms and get the terms inside the absolute value sign on one side. In this case, you would do it by adding 6 to both sides of the equation. Here’s how:

• |4x +2| – 6 = 8
• |4x +2| – 6 + 6 = 8 + 6
• |4x +2| = 14

Remove the absolute value and solve the equation. This is the first and easiest step. You’ll have to solve for x twice whenever you work with absolute value. Here’s how you do it the first time:

• 4x + 2 = 14
• 4x + 2 – 2 = 14 -2
• 4x = 12
• x = 3

Remove the absolute value and change the sign of the terms on the opposite side of the equal sign before you solve. Now, do it again, except set the first part of the equation equal to -14 instead of 14. Here’s how:

• 4x + 2 = -14
• 4x + 2 – 2 = -14 – 2
• 4x = -16
• 4x/4 = -16/4
• x = -4

Check your work. Now that you know that x = (3, -4), just plug both numbers back in to the equation to see that it works. Here’s how:

• (For x = 3):
  • |4x +2| – 6 = 8
  • |4(3) +2| – 6 = 8
  • |12 +2| – 6 = 8
  • |14| – 6 = 8
  • 14 – 6 = 8
  • 8 = 8
• (For x = -4):
  • |4x +2| – 6 = 8
  • |4(-4) +2| – 6 = 8
  • |-16 +2| – 6 = 8
  • |-14| – 6 = 8
  • 14 – 6 = 8
  • 8 = 8