What is an equation with one unknown?

Equations are algebraic expressions containing numbers and unknowns. It is important to differentiate these two groups: the numbers are fixed values while the unknowns, as their name indicates, represent unknown values (at least at the beginning), and in most cases we are asked to find out what this value is.
For example:

A1 - First-degree equations with one unknown

What do we do with the equations?

When we are given an exercise that contains an equation with an unknown, our goal is to solve the equation, that is, to find a solution to the equation. What does it mean to find the solution to an equation? The idea is to find the value of the unknown with the goal of making both sides of the equation equal.

When we have equations that have the same solution, they will be called equivalent equations.

When first degree equations include fractions, and the unknown is in the denominator, it is important to keep in mind the domain of the function


Practice Linear Equations

Examples with solutions for Linear Equations

Exercise #1

Solve for x x :

5x3=45 5x \cdot 3 = 45

Step-by-Step Solution

To solve the equation5x3=45 5x \cdot 3 = 45 , follow these steps:

1. First, identify the operation needed to solve forx x . In this case, we have a multiplication equation.

2. Therefore, we divide both sides of the equation by 15 (since 5×3=15 5 \times 3 = 15 ) to isolate x x :

x=4515 x = \frac{45}{15}

3. Calculate x x :

x=3 x = 3

Answer

x=3 x=3

Exercise #2

Solve for X:

2+x5=43 2 + x - 5 = 4 - 3

Step-by-Step Solution

To solve2+x5=43 2 + x - 5 = 4 - 3 , we first simplify both sides:

Left side:
25+x=3+x 2 - 5 + x = -3 + x

Right side:
43=1 4 - 3 = 1

Now the equation is 3+x=1 -3 + x = 1 .

Add 3 to both sides:
x=1+3 x = 1 + 3

So,x=4 x = 4 .

Answer

4

Exercise #3

Solve for X:

3x=106 3 - x = 10 - 6

Step-by-Step Solution

First, simplify the right side of the equation:
106=4 10 - 6 = 4
Hence, the equation becomes 3x=4 3 - x = 4 .
Subtract 3 from both sides to isolate x x :
3x3=43 3 - x - 3 = 4 - 3
This simplifies to:
x=1 -x=1
Divide by -1 to solve forx x :
x=1 x=-1
Therefore, the solution is x=1 x = 1 .

Answer

-1

Exercise #4

Solve for X:

3+x+1=62 3 + x + 1 = 6 - 2

Step-by-Step Solution

To solve 3+x+1=62 3 + x + 1 = 6 - 2 , we first simplify both sides:

Left side:
3+1+x=4+x 3 + 1 + x = 4 + x

Right side:
62=4 6 - 2 = 4

Now the equation is 4+x=4 4 + x = 4 .

Subtract 4 from both sides:
x=44 x = 4 - 4

So, x=0 x = 0 .

Answer

0

Exercise #5

Solve for X:

3+x2=73 3 + x - 2 = 7 - 3

Step-by-Step Solution

First, simplify both sides of the equation:

Left side: 3+x2=1+x 3 + x - 2 = 1 + x

Right side: 73=4 7 - 3 = 4

So the equation becomes:

1+x=4 1 + x = 4

Next, isolate x x by subtracting 1 from both sides:

1+x1=41 1 + x - 1 = 4 - 1

This simplifies to:

x=3 x = 3

Answer

3

Exercise #6

Solve for X:

5x=124 5 - x = 12 - 4

Step-by-Step Solution

First, simplify the right side of the equation:
124=8 12 - 4 = 8
Hence, the equation becomes 5x=8 5 - x = 8 .
Subtract 5 from both sides to isolate x x :
5x5=85 5 - x - 5 = 8 - 5
This simplifies to:
x=3 -x=3
Divide by -1 to solve for x x :
x=3 x=-3
Therefore, the solution is x=3 x=-3 .

Answer

-3

Exercise #7

Solve for X:

5+x3=2+1 5 + x - 3 = 2 + 1

Step-by-Step Solution

To solve 5+x3=2+1 5 + x - 3 = 2 + 1 , we first simplify both sides:

Left side:
53+x=2+x 5 - 3 + x = 2 + x

Right side:
2+1=3 2 + 1 = 3

Now the equation is 2+x=3 2 + x = 3 .

Subtract 2 from both sides:
x=32 x = 3 - 2

So, x=1 x = 1 .

Answer

1

Exercise #8

Solve for X:

6x=102 6 - x = 10 - 2

Step-by-Step Solution

To solve the equation 6x=102 6 - x = 10 - 2 , follow these steps:

  1. First, simplify both sides of the equation:

  2. On the right side, calculate 102=8 10 - 2 = 8 .

  3. The equation simplifies to 6x=8 6 - x = 8 .

  4. To isolate x, subtract 6 from both sides:

  5. 6x6=86 6 - x - 6 = 8 - 6

  6. This simplifies to x=2 -x = 2 .

  7. Multiply both sides by -1 to solve for x:

  8. x=2×1=2 x = -2 \times -1 = 2 .

  9. Since the problem requires only manipulation by transferring terms, the initial approach to the equation setup should lead to x = 4 as the solution before re-evaluation.

Therefore, the correct solution to the equation is x=2 x=2 .

Answer

2

Exercise #9

Solve for X:

7x=155 7 - x = 15 - 5

Step-by-Step Solution

First, simplify the right side of the equation:
155=10 15 - 5 = 10
Hence, the equation becomes 7x=10 7 - x = 10 .
Subtract 7 from both sides to isolate x x :
7x7=107 7 - x - 7 = 10 - 7
This simplifies to:
x=3 -x=3
Divide by -1 to solve forx x :
x=3 x=-3
Therefore, the solution is x=3 x=-3 .

Answer

-3

Exercise #10

Solve for X:

8x=113 8 - x = 11 - 3

Step-by-Step Solution

First, simplify the right side of the equation:
113=8 11 - 3 = 8
Hence, the equation becomes 8x=8 8 - x = 8 .
Subtract 8 from both sides to isolate x x :
8x8=88 8 - x - 8 = 8 - 8
This simplifies to:
x=0 -x=0
Divide by -1 to solve for x x :
x=0 x = 0
Therefore, the solution is x=0 x = 0 .

Answer

0

Exercise #11

Solve for X:

9x=167 9 - x = 16 - 7

Step-by-Step Solution

First, simplify the right side of the equation:
167=9 16 - 7 = 9
Hence, the equation becomes 9x=9 9 - x = 9 .
Since both sides are equal, x x must be 0 0 .
Therefore, the solution is x=0 x = 0 .

Answer

0

Exercise #12

Solve for X:

x3+5=82 x - 3 + 5 = 8 - 2

Step-by-Step Solution

First, simplify both sides of the equation:

Left side: x3+5=x+2 x - 3 + 5 = x + 2

Right side: 82=6 8 - 2 = 6

Now the equation is: x+2=6 x + 2 = 6

Subtract 2 from both sides to isolate x x :

x+22=62 x + 2 - 2 = 6 - 2

Simplifying gives:

x=4 x = 4

Answer

4

Exercise #13

Solve for X:

x+42=6+1 x + 4 - 2 = 6 + 1

Step-by-Step Solution

First, simplify both sides of the equation:

Left side: x+42=x+2 x + 4 - 2 = x + 2

Right side: 6+1=7 6 + 1 = 7

Now the equation is: x+2=7 x + 2 = 7

Subtract 2 from both sides to isolatex x :

x+22=72 x + 2 - 2 = 7 - 2

Simplifying gives:

x=5 x = 5

Answer

5

Exercise #14

Solve the equation:

6x2=24 6x \cdot 2 = 24

Step-by-Step Solution

To solve the equation 6x2=24 6x \cdot 2 = 24 , follow these steps:

1. First, identify the operation involved. In this case, it is multiplication.

2. Divide both sides of the equation by 12 (since 6×2=12 6 \times 2 = 12 ) to isolate x x :

x=2412 x = \frac{24}{12}

3. Calculate x x :

x=2 x = 2

Answer

x=2 x=2

Exercise #15

Solve the equation

20:4x=5 20:4x=5

Video Solution

Step-by-Step Solution

To solve the exercise, we first rewrite the entire division as a fraction:

204x=5 \frac{20}{4x}=5

Actually, we didn't have to do this step, but it's more convenient for the rest of the process.

To get rid of the fraction, we multiply both sides of the equation by the denominator, 4X.

20=5*4X

20=20X

Now we can reduce both sides of the equation by 20 and we will arrive at the result of:

X=1

Answer

x=1 x=1