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How to Solve Equations Using Substitution


Sometimes expressions are especially complex, and a clever trick to solve them is to simplify them by using substitution. When you substitute, you exchange an expression with a variable. The variables often used are u, v, w, and z.

Theory

Substitution

Substitution is when you substitute a variable for part of an expression:

1.
You substitute a variable, such as u, for a complex factor in the expression, then calculate with the variable in place.
2.
Once your calculation is complete, you substitute the original factor back into the result for the variable u, and solve.

Example 1

Solve the equation 22x 2x+1 + 1 = 0

This one looks a bit weird, but with a little help from the rules of calculating with powers it gets easier. 22x 2x+1 + 1 = 0

Transform the equation so you can use substitution: (2x) 2 2 2x + 1 = 0

You can see that 2x = u gives us an ordinary square equation: u2 2 u + 1 = 0 u2 2u + 1 = 0

You can see that this is a quadratic equation which is solved either by using the quadratic formula, or by factorizing. You can therefore find the uvalues: u2 2u + 1 = 0 (u 1) (u 1) = 0 (u 1) 2 = 0 u 1 = 0 u = 1

Put the values for u in to 2x = u and solve for x: 2x = 1 log 2x = log 1 x log 2 = 0 (log 1 = 0) x = 0 log 2 = 0

Example 2

Solve the equation x4 5x2 + 6 = 0

To solve an equation like this, it helps to recognize that x4 = (x2) 2. You can substitute and solve it as a normal quadratic equation: x4 5x2 + 6 = 0 (x2) 2 5x2 + 6 = 0

Let x2 = u and substitute: u2 5u + 6 = 0

You can solve this equation either with the quadratic formula, or with inspection: u = 5 ± (5 ) 2 4 1 6 2 1 = 5 ±25 24 2 = 5 ±1 2 = 5 ± 1 2

Now you split the two equations, one with a positive root in the numerator and one with a negative root in the numerator: u1 = 5 + 1 2 u2 = 5 1 2 = 6 2 = 4 2 = 3 = 2

At the end you put these values back in the substitution 2x = u to find the values for x: x2 = 3 x2 = 2 x = ±3 x = ±2

The answer you get is then x1 = 3, x2 = 3, x3 = 2 and x4 = 2.

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