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An equation is an expression that is equal to another expression.
Example 15.1
Examples of equations:
a) \(x-4=6\)
b) \(5 x-6=4 x+2\)
c) \(6 x-30=0\)
d) \(x^{2}+3 x-4=0\)
e) \(3 x^{2}-2 x=-1\)
f) \(x^{3}+x^{2}+x+1=0\)
g) \(2 x-5>3\) is not an equation. It is an inequality and will be discussed in chapter 21
A solution of an equation is any value of the variable that satisfies the equality, that is, it makes the Left Hand Side (LHS) and the Right Hand Side (RHS) of the equation the same value.
To solve an equation is to find the solution(s) for that equation. The method to solve an equation depends on the kind of equation at hand. We will study how to:
- solve linear equations in chapters 16 and 17
- solve quadratic equations in chapter 20
Example 15.2
Solutions of equations:
a) A solution for \(x-4=6\) is \(x=10\) because the LHS evaluated at \(x=10\) is \(10-4=6\) which is equal to the \(\mathrm{RHS}\).
b) A solution for \(5 x-6=4 x+2\) is \(x=8\) because the LHS evaluated at \(x=8\) is \(5(8)-6=40-6=34\) and the RHS evaluated at \(x=8\) is \(4 x+2=4(8)+2=32+2=34,\) and they are equal!
So, given a value of \(x\), we can check if it is a solution or not by evaluating simultaneously the LHS and RHS of an equation. If they are equal, then the value is a solution. If they are not equal, then the value is not a solution.
Example 15.3
a) Is \(x=2\) a solution of the equation
\[-4 x+8+x=5-2 x+1\nonumber\] ?
The LHS evaluated at \(x=2\) is \(-4(2)+8+2=-8+8+2=2\).
The RHS evaluated at \(x=2\) is \(5-2(2)+1=5-4+1=5+(-4)+1=1+1=2\).
Since they are equal, then we say that \(x=2\) is a solution for the given equation.
b) Is \(x=-1\) a solution of the equation
\[x^{2}+4 x=-3 x+8\nonumber\] ?
The LHS evaluated at \(x=-1\) is \((-1)^{2}+4(-1)=1+(-4)=-3\).
The RHS evaluated at \(x=-1\) is \(-3(-1)+8=3+8=11\).
Since \(-3 \neq 11\), then we say that \(x=-1\) is not a solution for the given equation.
c) Is \(x=-2\) a solution to
\[x^{2}-2 x+1=3 x^{2}+2 x+1\nonumber\] ?
The LHS evaluated at \(x=-2\) is \((-2)^{2}-2(-2)+1=4+4+1=9\).
The RHS evaluated at \(x=-2\) is \(3(-2)^{2}+2(-2)+1=3 \cdot 4-4+1=12-4+1=9 .\) since \(9=9,\) the \(L H S=R H S\) and \(x=-2\) is a solution to the equation.
Exit Problem
Check: Is \(x=-6\) a solution of the equation
\[10+10 x=13+6 x+1 ?\nonumber\]
