Equations are statements of equality.
x + 1 = 2 is a statement of equality between x + 1 and 2.
The statement is:
For what value of x, will x + 1 become equal to 2 ?
Answer: for x = 1.
Replacing x by 1 makes x + 1 equal to 2,
i.e. 1 + 1 = 2.
We say, x = 1 is solution of the equation:
x + 1 = 2.
Replacing x by 1 is called substitution
In the equation x + 1 = 2:
x + 1 is said to be in the Left Hand Side, in short LHS, and 2, in the Right Hand Side, abbreviated as RHS
x is called variable, i,e what changes or varies.
The values of x for which the LHS and the RHS become equal are called Solution Set of the Equation.
We also say solution of x satisfies the equation.
To satisfy an equation means the values of x which make LHS and RHS equal.
“To solve for x” is to find the solution of an equation.
An equation in which the highest exponent of the variable is 1 is called a linear equation.
In x + 1 = 2, the highest exponent of the variable x is 1.
So, we say x + 1 = 2 is a linear equation in variable x.
Other examples of linear equations are:
x + 3 = 5, in which the highest exponent of the variable x is 1
p + 9 = 2, in which the highest exponent of the variable p is 1
a + b = 1, in which the highest exponents of the two variables a and b are 1
Equations in which the highest exponent of the variable is 2 are called quadratic equations .
Examples of quadratic equations are:
x2 + 2x + 1 = 0, in which the highest exponent of the variable x is 2
x2 – 4 = 0, in which the highest exponent of the variable x is 2
y2 – 4ax = 0, in which the highest exponent of the variable y is 2
6p2 – p - 1 =0, in which the highest exponent of the variable p is 2.
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Solve for x:
x + 2 = 3
Transpose 2 to RHS, and reverse its sign.
x = 3 – 2 = 1, so x = 1 is the solution.
Solving two equations x + y = 3 and x – 2y = 1 for the two different variables x and y so that the solution satisfies both the given equations are called simultaneous equations.
Solve the equations:
x – y = 2 and x + y = 1.
From the first equation, transpose y to the right by inversing its sign:
x = y + 2,
substitute this value of x, i.e. y + 2 in x in the second equation:
y + 2 + y = 1,
so that, 2y = 1 – 2, 2y = -1, y = -1/2
now, substitute y = -1/2 in any of the two equations to solve for x:
(let us substitute in the first equation)
x – (-1/2) = 2, i.e. x + ½ = 2, so that x = 2 – ½ = 3/2
so, the solution, i.e. values of the two variables x and y, for the pair of equations is x = 3/2 and y = -1/2
Solve the system of equations:
2x – 3y = 4 and 3x – 4y = 8.
Let us eliminate y by making coefficients of y in the two equations equal.
Multiply first equation with 4, the coefficient of y (disregard the minus sign) in the 2nd equation, and the second with 3, the coefficient of y (here too disregard the minus sign) in the 1st equation as below:
4 × (2x – 3y = 4), i.e. 8x – 12y = 16, and
3 × (3x – 4y = 8), i.e. 9x – 12y = 24.
Now subtract the new 2nd equation from the new 1st, and you will see the y-term gone.
[8x – 12y – 16] – [9x – 12y – 24] = 0
i.e. –x + 8 = 0, i.e. –x = - 8 , i.e. x = 8.
Now, substitute x = 8 in any of the two equations, say in the 1st (the original) to solve for y:
2 × 8 – 3y = 4,
i.e. -3y = 4 – 16,
i.e. -3y = -12,
3y = 12, i.e. y = 12/3 = 4.