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INAIO Guide Stage 1 2Algebra and Equations

2.2 Linear equations and systems

Keep an equation balanced, solve for one unknown, and find values that satisfy two equations at once.

An expression tells us how to calculate a quantity. An equation says that two quantities are equal and asks when that statement is true. A system asks for values that make several equations true at the same time. These ideas appear whenever two descriptions of the same situation must agree: a total price, a measured distance, or two counts with a shared total.

This is foundational algebra. Review the expression rules in Module 2.1, then use OpenStax's one-variable equations before its two-variable systems. You need only substitution and arithmetic to start.

A basic idea: preserve the balance

Consider 2x+3=112x+3=11. The left and right sides have the same value only for certain xx. Subtracting 33 from both sides leaves 2x=82x=8; dividing both sides by 22 gives x=4x=4. Checking in the original equation gives 2(4)+3=112(4)+3=11.

An equation is like a balanced scale: adding or subtracting the same quantity on both sides preserves its solutions, as does multiplying or dividing by the same nonzero number. Dividing by a variable requires care because that variable might be zero. For example, dividing x2=xx^2=x by xx would lose the valid solution x=0x=0.

A quick tour of equations and systems

One unknown

A linear equation in one variable can be rearranged to ax=bax=b. If a≠0a\ne0, it has the single solution x=b/ax=b/a. Simplify brackets, gather the variable terms on one side, gather constants on the other, then check the result in the original equation.

If rearrangement instead produces 0=50=5, there is no solution: no value can make a false statement true. If it produces 0=00=0, every allowed value works. For instance, 2(x+1)=2x+22(x+1)=2x+2 is true for all real xx.

Two unknowns at once

In a system such as

x+y=5,2x−y=1,x+y=5,\qquad 2x-y=1,

a solution is a pair (x,y)(x,y) that satisfies both equations. Adding the equations removes yy: 3x=63x=6, so x=2x=2 and then y=3y=3. Substitution checks both: 2+3=52+3=5 and 2(2)−3=12(2)-3=1.

Two common methods are:

Method What you do Good time to use it
Substitution Express one variable using the other, then put it into the second equation. One variable is already isolated or easy to isolate.
Elimination Add or subtract equations after matching coefficients, removing one variable. Coefficients already match or can be matched easily.

Graphically, each two-variable linear equation is a line. The two graphs can cross once, run parallel and never meet, or coincide. Their system then has one, zero, or infinitely many solutions, respectively. The equations x+y=5x+y=5 and 2x−y=12x-y=1 cross at (2,3)(2,3).

Coordinate graph with the lines x plus y equals 5 and 2x minus y equals 1 crossing at the labelled point (2,3).Coordinate graph with the lines x plus y equals 5 and 2x minus y equals 1 crossing at the labelled point (2,3).
The intersection gives the pair that satisfies both equations, not just one of them.

For more than two unknowns, the same idea continues: every equation must hold. Matrix methods in Module 11.4 organise larger systems.

Application 1: an equation with brackets

Solve 3(2x−1)=2x+133(2x-1)=2x+13.

  1. Expand the left side. 6x−3=2x+136x-3=2x+13.
  2. Collect terms. Subtract 2x2x from both sides, then add 33: 4x=164x=16.
  3. Solve and check. x=4x=4. The original sides are 3(8−1)=213(8-1)=21 and 2(4)+13=212(4)+13=21.

Checking in the original form catches a missed term when expanding.

Application 2: adult and child tickets

A group buys 1717 tickets. Adult tickets cost 88 units, child tickets cost 55 units, and the total paid is 109109 units. How many of each were bought?

  1. Name the unknowns. Let aa be adult tickets and cc child tickets. Then a+c=17a+c=17 and 8a+5c=1098a+5c=109.
  2. Substitute. From the count equation, c=17−ac=17-a. Thus 8a+5(17−a)=1098a+5(17-a)=109, or 3a=243a=24.
  3. Find and check both counts. a=8a=8 and c=9c=9. They add to 1717, and 8(8)+5(9)=64+45=1098(8)+5(9)=64+45=109.

The count and cost equations tell us different things. Both are needed to determine the pair.

Resources

SourceTitleWhy read it
OpenStaxLinear equations in one variableStart with balancing and checking one-variable equations.
OpenStaxSystems of linear equations: two variablesCompare graphing, substitution, and elimination.

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