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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Solving Systems of Equations Using Substitution

Two algebraic methods for solving systems of equations are addition and substitution.

A â€œsystem of equationsâ€ is defined by the set of all of the lines that intersect at the point (h, k) and the solution which can be determined by the equations of two different lines. The simplest of these are { x = h, y = k } which is called the â€œsolution setâ€ of the system of equations. The solution is also defined by the point of intersection: { (h, k) }.

In any system of equations (family of lines) the following are true statements.

1. A linear equation may be multiplied by a real number and form another linear equation through the same point. (a coincident line).

2. Two intersecting linear equations may be added to form the equation of another line through the same point on a graph.

3. One equation can be â€œsolved for one of the variablesâ€ and that expression can then be substituted in another equation in place of that variable.

NOTE: [Generally, use only for variables with coefficient of Â±1.]

## Substitution Method

Unless one of the equations is already solved for one of the variables in terms of the other, the Substitution Method is not generally as desirable as the Addition Method.

Example 1:

Since the second equation is already solved for y we can use the expression on the right side as a replacement for y in the first equation, which gives us an equation in x alone.

Solve:

 x + (2x - 1) = 11 Replacement for y 3x - 1 = 11 Distributive Property 3x = 12 Add Opps x = 4 Multiply Recip

Replace x = 4 in the second equation y = 2( 4 ) − 1 or y = 7.

Check: Substitute both in equation one: ( 4 ) + ( 7 ) = 11 The solution point: { ( 4, 7) }

Example 2:

Since the first equation is already solved for y we can use the expression on the right side as a replacement for y in the second equation, which gives us an equation in x alone.

Substitute the expression for y in the second equation: 3x − 2(-x + 8) = − 1

Solve this equation for x:

 3x + 2x − 16 = − 1 Distributive property 5x = 15 Add Opps x = 3 Multiply Recip

Replace x = 3 in equation (1): y = − ( 3 ) + 8 finds y = 5 or solution point: { ( 3, 5) }

Always check in the other equation: 3(3) − 2(5) = − 1

Never consider using the substitution method if all of the coefficients in the system are different than Â±1. If none are equal to Â±1 you may have fractions to contend with and thatâ€™s just not necessary. Some simple systems do have fractions as solutions.

In fact, if you encounter a system of equations where the coefficients are fractions, it is usually better to multiply first by the LCM then proceed to solve the system.