One step algebra equation
Introduction:
In mathematics, an algebraic equation, also called polynomial equation is an equation of the form P = Q, where P and Q are polynomials over that field. Two equations are equivalent if they have the same set of solutions. In particular the equation P = Q is equivalent with P − Q = 0. For example y4 + (xy/2) = (x3/3) – xy2 + y2 – (1/7) is an algebraic equation over the rationals. Let us see about one step algebra equation.
One step algebra equation:
In one step algebra equation, by combining variables and constants we make algebraic expressions. The basic operations of addition, subtraction, multiplication and division are used for combining the variables and constants. By combining the variables and constants we get the algebraic expressions like 7x + 32, 16y – 10. The expression 7x + 32 is obtained from the variable x, first by multiplying "x" by the constant 7 and then adding the constant 32 to the product.
Similarly, 16y – 10 is obtained by first multiplying y by 16 and then subtracting 10 from the product.
Look at how the following expressions are obtained:
x2, 5y 2, 5x2 – 12, xy, 7xy + 45
The expression x2 is obtained by multiplying the variable x by itself; x × x = x2. Just as 5 × 5 is written as 52, we write x × x = x2. In the same manner, we can write x × x × x = x3 commonly, x3 is read as ‘x cubed’. We can realize that x3 may also be read as x raised to the power 3. x, x2, x3 ... are all algebraic expressions obtained from x.
Problems of one step algebra equation:
Ex 1: Solve for x in the following equation. x - 6 = 20
Sol: Add 6 to both sides of the equation:
x = 26
Ans: x = 14
Ex 2: Solve for x in the following equation. 3x - 4 = 11
Sol: Add 4 to both sides of the equation:
3x = 15
Divide both sides by 3: x = 5
Ans: x = 5
Ex 3: Find the x in the following equation 6 x - 6 = 2 x - 8
Sol: Subtract 2x from both sides of the equation:
4x – 6 = - 8
Add 6 to both sides of the equation:
4x = -2
Divide both sides by 4: x = -1/2
Ans: x = - `1/2`
Ex 4: Find X for following algebraic expression: x + 10 = 20 and x – 3 = 10.
Sol:
Given, X + 10 = 20
Let us subtract 10 on both sides we get,
x + 10 -10 =20 - 10
x = 10
Given, X -3 = 10
Let us add 3 on both sides, we get
x - 3 + 3 = 10 +3
x = 13
Ans: x=10 and x=13.
Ex 2: Find n for following algebraic expression: 4n = 20 and 2n + 3 = 5. Using basic format of algebric expression.
Sol:
Given, 4n = 20
Let us divide by 4 on both sides, we get
`(4n)/4` = `20/4`
n = 5.
Given, 2n + 3 = 5
Let us subtract by 3 on both sides, we get
2n + 3 -3 = 5 - 3
2n = 2
Let us divide by 2 on both sides, we get
`( 2n)/2 = 2/2`
n = 1.
Ans: n = 5 and n = 1.
In mathematics, an algebraic equation, also called polynomial equation is an equation of the form P = Q, where P and Q are polynomials over that field. Two equations are equivalent if they have the same set of solutions. In particular the equation P = Q is equivalent with P − Q = 0. For example y4 + (xy/2) = (x3/3) – xy2 + y2 – (1/7) is an algebraic equation over the rationals. Let us see about one step algebra equation.
One step algebra equation:
In one step algebra equation, by combining variables and constants we make algebraic expressions. The basic operations of addition, subtraction, multiplication and division are used for combining the variables and constants. By combining the variables and constants we get the algebraic expressions like 7x + 32, 16y – 10. The expression 7x + 32 is obtained from the variable x, first by multiplying "x" by the constant 7 and then adding the constant 32 to the product.
Similarly, 16y – 10 is obtained by first multiplying y by 16 and then subtracting 10 from the product.
Look at how the following expressions are obtained:
x2, 5y 2, 5x2 – 12, xy, 7xy + 45
The expression x2 is obtained by multiplying the variable x by itself; x × x = x2. Just as 5 × 5 is written as 52, we write x × x = x2. In the same manner, we can write x × x × x = x3 commonly, x3 is read as ‘x cubed’. We can realize that x3 may also be read as x raised to the power 3. x, x2, x3 ... are all algebraic expressions obtained from x.
Problems of one step algebra equation:
Ex 1: Solve for x in the following equation. x - 6 = 20
Sol: Add 6 to both sides of the equation:
x = 26
Ans: x = 14
Ex 2: Solve for x in the following equation. 3x - 4 = 11
Sol: Add 4 to both sides of the equation:
3x = 15
Divide both sides by 3: x = 5
Ans: x = 5
Ex 3: Find the x in the following equation 6 x - 6 = 2 x - 8
Sol: Subtract 2x from both sides of the equation:
4x – 6 = - 8
Add 6 to both sides of the equation:
4x = -2
Divide both sides by 4: x = -1/2
Ans: x = - `1/2`
Ex 4: Find X for following algebraic expression: x + 10 = 20 and x – 3 = 10.
Sol:
Given, X + 10 = 20
Let us subtract 10 on both sides we get,
x + 10 -10 =20 - 10
x = 10
Given, X -3 = 10
Let us add 3 on both sides, we get
x - 3 + 3 = 10 +3
x = 13
Ans: x=10 and x=13.
Ex 2: Find n for following algebraic expression: 4n = 20 and 2n + 3 = 5. Using basic format of algebric expression.
Sol:
Given, 4n = 20
Let us divide by 4 on both sides, we get
`(4n)/4` = `20/4`
n = 5.
Given, 2n + 3 = 5
Let us subtract by 3 on both sides, we get
2n + 3 -3 = 5 - 3
2n = 2
Let us divide by 2 on both sides, we get
`( 2n)/2 = 2/2`
n = 1.
Ans: n = 5 and n = 1.