GRE Algebra | Solving linear Equations (original) (raw)
Last Updated : 15 Apr, 2024
In mathematics, a **linear equation is an equation that may be put in the form,
a1x1 +........... + anxn + b = 0
Where x1, xn are the variables, and a1, an, b are the coefficients, which are often real numbers.
To solve a linear equation means to find the values of all the variables that make the equation satisfy. Two equations said to be **equivalent equation when both of them have same solution.
For example:
y + 4 = 16 and and 3y + 12 = 48 both are equivalent equation because both are true for (y = 12).
**Rules for producing equivalent equations:
- When same constant added to or subtracted from both sides of an equation then equality is preserved and new equation is equivalent to the original equation.
- When the equation is multiplied or divided by the same non-zero constant on both sides, the equality is preserved and new equation is equivalent to the original equation.
**Linear equations in one variable:
A linear equation in one variable, can be written in form,
ax + b = c
where a, b are real numbers. Linear equation can be solved by combining all the like terms on each side of the equation. Then use all the rules of producing similar equivalent equations.
**Example-1: Solve 15x - 2x - 9 = 4x + 3(x - 5)
**Solution:
13x - 9 = 4x + 3x - 15
13x - 9 = 7x - 15 (like terms combined)
13x - 9 -7x = 7x -15 - 7x (-7x) added to both sides
6x - 9 = -15
6x - 9 + 9 = -15 + 9 (9) is added to both sides
6x = -6
6x/6 = -6/6 both sides divided by 6
x = -1
**Linear equations in two variables:
A linear equation in two variables x and y, can be written in the form,
ax + by + c = 0
where a, b and c are real numbers but a and b not zero at the same time. Two variable equation can be solved by two methods:
- Substitution method
- Elimination method
**Example-2: Solve using **substitution method 2x + 2y = 4 and 4x – 6y = 18
**Solution: We can express x in terms of y in first equation as,
x = 2 - y
Then, substitute the (2 - y) for x in the second equation to find the value of y.
4(2 - y) - 6y = 18
8 - 4y - 6y = 18
-10y = 10
**y = -1
Substitute value of y in either equation to find the value of x.
2x + 2(-1) = 4
2x = 6
**x = 3
**Example-3: Solve using **elimination method 2x + 2y = 4 and 4x – 6y = 18
**Solution: Compare the coefficients of x and y and make equal for one variable. Then subtract or add both the equations so that one variable can be eliminated. Multiply first equation by 2 and subtract the second equation from first.
4x + 4y = 8 ..........(1)
4x - 6y = 18 .........(2)
10y = -10 **y = -1
Hence, put value of y in first equation,
2x + 2(-1) = 4
2x = 6
**x = 3