Help solving linear equations

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Intermediate Algebra
Tutorial 7: Linear Equations in One Variable


WTAMU> Virtual Math Lab > Intermediate Algebra  



deskLearning Objectives


 
After completing this tutorial, you should be able to:
  1. Know what a linear equation is.
  2. Know if a value is a solution or not.
  3. Use the addition, subtraction, multiplication, and division properties of equalities to solve linear equations.
  4. Know when an equation has no solution.
  5. Know when an equation has all real numbers as a solution. 




desk Introduction



This is where we start getting into the heart of what algebra is about - solving equations.  In this tutorial we will be looking specifically at linear equations and their solutions.  We will start off slow and solve equations that use only one property to make sure you have the individual concepts down.  Then we will pick up the pace and mix 'em up where you need to use several properties and steps to get the job done. 

Equations can be used to help us solve a variety of problems. In later tutorials, we will put them to use to solve word problems.  Then you can answer those tricky math questions.

 

 

desk Tutorial


 
 
 
 
Equation

Two expressions set equal to each other


 
Linear Equation

An equation that can be written in the form 
ax + b = c
where a, b, and c are constants


 
 

The following is an example of a linear equation: 3x - 4 = 5


 
 
 
Solution

A value, such that, when you replace the variable with it,
it makes the equation true.

(the left side comes out equal to the right side)


 
Solution Set

Set of all solutions


 
 
notebook Example 1: Determine if any of the following values for x are solutions to the given equation.   
3x - 4 = 5;   x = 3, 5.

 

Checking 3
3x - 4 = 5
3(3) - 4 = 5 
9 - 4 = 5 
5 = 5 
True          3 is a solution

Checking 5
3x - 4 = 5
3(5) - 4 = 5 
15 - 4 = 5 
11 = 5 
False         5 is not a solution


 
 
 
Solving a Linear Equation 
in General

Get the variable you are solving for alone on one side and everything else on the other side using INVERSE operations.


 

The following will give us the tools that we need to solve linear equations.


 
Addition and Subtraction Properties of Equality

If a = b, then a + c = b + c

If a = b, then a - c = b - c


 

In other words, if two expressions are equal to each other and you add or subtract the exact same thing to both sides, the two sides will remain equal.

Note that addition and subtraction are inverse operations of each other.  For example, if you have a number that is being added that you need to move to the other side of the equation, then you would subtract it from both sides of that equation. 
 
 

notebook Example 2:  Solve for the variable.     x - 5 = 2.


 

x - 5 = 2
x - 5 + 5 = 2 + 5
x = 7

*Inverse of sub. 5 is add. 5 


 

Note that if you put 7 back in for x in the original problem you will see that 7 is the solution to our problem.


 
 
 
notebook Example 3: Solve for the variable.      y + 4 = -7.

 

y + 4 = -7
y + 4 - 4 = -7 - 4
y = -11

*Inverse of add. 4 is sub. 4


 

Note that if you put -11 back in for y in the original problem you will see that -11 is the solution we are looking for.


 
 
 
Multiplication and Division Properties of Equality

If a = b, then a(c) = b(c)

If a = b, then a/c = b/c where c is not equal to 0.


 

In other words, if two expressions are equal to each other and you multiply or divide (except for 0) the exact same constant to both sides, the two sides will remain equal. 

Note that multiplication and division are inverse operations of each other.  For example, if you have a number that is being multiplied that you need to move to the other side of the equation, then you would divide it from both sides of that equation. 

Note, for multiplication and division, it is not guaranteed that if you multiply by the variable you are solving for that the two sides are going to be equal.  But is guaranteed that the two sides are going to be equal if you are multiplying or dividing by a constant or another variable that you are not solving for.  We will talk more about this in a later tutorial.  For this tutorial just note you can use this property with constants and variables you are not solving for. 
 

notebook Example 4:  Solve for the variable.     x/2 = 5.


 
example 4

*Inverse of div. by 2 is mult.  by 2


 

If you put 10 back in for x in the original problem, you will see that 10 is the solution we are looking for.


 
 
notebook Example 5:   Solve for the variable.    5x = 7.

 
example 5

*Inverse of mult. by 5 is div. by 5


 

If you put 7/5 back in for x in the original problem, you will see that 7/5 is the solution we are looking for.


 
 

The examples above were using only one property at a time to help you understand the different properties that we use to solve equations.  However, most times, we have to use several properties to get the job done.  The following is a strategy that you can use to help you solve linear equations that are a little bit more involved.


 
 
Strategy for Solving a Linear Equation

 

Note that your teacher or the book you are using may have worded these steps a little differently than I do, but it all boils down to the same concept -   get your variable on one side and everything else on the other using inverse operations.
 

Step 1: Simplify each side, if needed.
 

This would involve things like removing ( ), removing fractions, adding like terms, etc. 

To remove ( ):  Just use the distributive property found in Tutorial 5: Properties of Real Numbers.

To remove fractions: Since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of your fractions.

 

Step 2: Use Add./Sub. Properties to move the variable term to one side and all other terms to the other side. 
 

Step 3: Use Mult./Div. Properties to remove any values that are in front of the variable.
 

Step 4:  Check your answer.
 

I find this is the quickest and easiest way to approach linear equations.
 
 
 

notebook Example 6:  Solve for the variable.     10 - 3x = 7.


 
example 6

*Inverse of add. 10 is sub. 10
 
 

*Inverse of mult. by -3 is div. by -3


 

Be careful going from line 4 to line 5.  Yes, there is a negative sign. But, the operation between the -3 and x is multiplication not subtraction.  So if you were to add 3 to both sides you would have ended up with -3x + 3 instead of the desired x

If you put 1 back in for x in the original problem you will see that 1 is the solution we are looking for.


 
 
 
notebook Example 7:  Solve for the variable.     2(x + 5) - 7 = 3(x - 2).

 
example 7

*Remove ( ) by using dist. prop.
 

*Get all x terms on one side

*Inverse of add. 3 is sub. 3
 
 

*Inverse of mult. by -1 is div. by -1
 


 

If you put 9 back in for x in the original problem you will see that 9 is the solution we are looking for.


 
 
notebook Example 8:   Solve for the variable: example 8a.

 
example 8b

*To get rid of the fractions, 
mult. both sides by the LCD of 4

*Get all the x terms on one side
 

*Inverse of add. 2 is sub. 2
 
 

*Inverse of mult. by -3 is div. by -3

 


 

If you put 4/3 back in for x in the original problem you will see that 4/3 is the solution we are looking for.


 
 
Contradiction

A contradiction is an equation with one variable that has no solution.


 
 
notebook Example 9:   Solve for the variable.   4x- 1 = 4(x+ 3).

 
example 9


*Remove ( ) by using dist. prop.

*Get all the x terms on one side
 


 

Where did our variable, x, go???  It disappeared on us.  Also note how we ended up with a FALSE statement, -1 is not equal to 12.  This does not mean that x = 12 or x = -1. 

Whenever your variable drops out AND you end up with a false statement, then after all of your hard work, there is NO SOLUTION.

So, the answer is no solution.


 
 
 
Identity

An identity is an equation with one variable
that has all real numbers as a solution.


 
 
notebook Example 10:   Solve for the variable.   5x+ 10 = 5(x+ 2).

 
example 10


*Remove ( ) by using dist. prop.

*Get all the x terms on one side
 


 

This time when our variable dropped out, we ended up with a TRUE statement.  Whenever that happens your answer is ALL REAL NUMBERS.

So, the answer is all real numbers.


 

 
desk Practice Problems
  

These are practice problems to help bring you to the next level.  It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it.  Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument.  In fact there is no such thing as too much practice.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that  problem.  At the link you will find the answer as well as any steps that went into finding that answer.

 

pencil Practice Problems 1a - 1e: Solve for the variable.

 

 

 


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WTAMU> Virtual Math Lab > Intermediate Algebra


Last revised on July 1, 2011 by Kim Seward.
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Current time:0:00Total duration:7:27

Old school equations with Sal

Video transcript

Sours: https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq/alg-old-school-equations/v/algebra-linear-equations-1
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Linear Equations

6.2  Solving Linear Equations

  Equations of the form ax+b=0 are called linear equations in the variable x. In this section we will be concerned with the problem of solving linear equations, and equations that reduce to linear equations.

  We define two equations as equivalent if they have the same solution set. The following two operations on an equation always result in a new equation which is equivalent to the original one. These operations, sometimes called elementary transformations, are:

  T.1 The same expression representing a real number may be added to both sides of an equation.

  T.2 The same expression representing a nonzero real number may be multiplied into both sides of an equation.

  Using these operations we may transform an equation whose solution set is not obvious through a series of equivalent equations to an equation that has an obvious solution set.

Example 1.  Solve the equation

  (a) 2x-3=4+x

  Add -x to both sides to obtain

       -x+2x-3=-x+4+x  (T.1)

  or  x-3=4

  Add 3 to both sides to obtain

       x-3+3=4+3  (T.1)

  or  x=7

  Since 2x-3=4+x is equivalent to x-3=4, which, in turn, is equivalent to x=7, whose solution set is obviously {7}, we know that the solution set of (a) is {7}.

Let’s see how our Linear equation solver solves this and similar problems. Click on "Solve Similar" button to see more examples. 

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Example 2.  Solve the equation

  (b) 1/2x+2/3=5/2x-1

  Add -(1/2)x to both sides to obtain

       2/3=5/2x-1/2x-1  (T.1)

  or  2/3=2x-1

  Add 1 to both sides to obtain

       1+2/3=2x  (T.1)

  or  5/3=2x

  Multiply both sides by 1/2 to obtain

       5/6=x  (T.2)

  Thus, the solution set of (b) is {5/6}.

  Every linear equation can be solved in the same way as in the above examples. In fact, let us consider the general linear equation

  ax+b=0

  Add -b to both sides to obtain

  ax=-b

  Multiply both sides by 1/a to obtain

  x=-(b/a)

  if a a!=0. The general linear equation, therefore, has as its solution set {b/a}, if a!=0. Thus each linear equation has at most one solution.

  The next two examples are of equations that reduce to linear equations.

Example 3.  Solve the equation

  23+4y(5y+4)=9+10y(2y+3)

  We expand both sides to obtain

  23+20y^2+16y=9+20y^2+30y

  Add -20y^2 to both sides to obtain

  23+16y=9+30y

  We now solve as in the previous examples.

  23+16y=9+30y

  23-9=30y-16y

  14=14y

  y=1

  Thus the solution set is {1}.

Let’s see how our step by step math solver solves this and similar problems. Click on "Solve Similar" button to see more examples. 

Solve similiar problemEnter your own problem

Example 4.  Solve the equation

  (c) (2x)/(x-1)=2/(x-1)+1

  The replacement set of (c) is all real numbers except 1. Assuming that x!=1, we multiply both sides of (c) by x-1 to obtain

  (d) 2x=2+x-1,x!=1

  Solving the equation 2x =2+x- 1, we obtain 1 as the only solution Since 1 is not in the replacement of (d), (d) has no solution. Furthermore, (c) is equivalent to (d), therefore (c) has no solution.

6.3  Solving Literal Equations

  An equation containing more than one variable, or containing symbols representing constants such as a,b, and c, can be solved for one of the symbols in terms of the remaining symbols by applications of the operations T.1 and T.2 in the preceding section. The student will encounter such problems in other courses.

Example 1.  Solve cx-3a=b for x.

  Add 3a to both sides.

  cx=b+3a

  Multiply both sides by 1/c.

  x=(b+3a)/c

  This last equation expresses x in terms of the other symbols.

Example 2.  Solve 3ay-2b=2cy for y.

  Add 2b to both sides.

  3ay=2cy+2b

  Add -2cy to both sides.

  3ay-2cy=2b

  Factor out y.

  (3a-2c)y=2b

  Multiply both sides by 1/((3a-2c))

  y=(2b)/(3a-2c)

Example 3.  Solve a/x+b/(2x)=c for x.

  Multiply both sides by 2x.

  2a+b=2cx

  2cx=2a+b

  Multiply by 1/(2c).

  x=(2a+b)/(2c)

  We conclude this section by including two more examples similar to those that the student may encounter in other areas.

Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples. 

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Example 4.  Solve A=P(1+rt) for r.

  Apply the distributive law.

  A=P+Prt

  Add -P to both sides.

  A-P=Prt

  Multiply both sides by 1/(Pt).

  (A-P)/(Pt)=r

Example 5.  Solve 1/R=1/r_1+1/r_2 for r_1.

  Add the two terms on the right—hand side.

  1/(R)=(r_2+r_1)/(r_1r_2)

  Multiply by Rr_1r_2.

  r_1r_2=R(r_2+r_1)

  r_1r_2=Rr_2+Rr_1

  Add -Rr_1 to both sides.

  r_1r_2-Rr_1=Rr_2

  Factor out r_1.

  r_1(r_2-R)=Rr_2

  Multiply by 1/(r_2-R).

  r_1=(Rr_2)/(r_2-R)

6.4  Solving Statement Problems

  One of the fundamental applications of algebra is solving problems that have been stated in words. A statement problem is a word description of a situation that involves both known and unknown quantities. In this section each problem will be solvable by means of one equation involving one unknown.

  Our problem is to choose the unknown and to determine the equation that it must satisfy. Although there is no single approach to all of the problems, the following suggestions are sometimes helpful:

  1. Read the problem carefully until the situation is thoroughly understood.

  2. Determine what quantities are asked for, then choose the one that seems to be the best to use as the unknown.

  3. Establish the relationship between the unknown and the other quantities in the problem.

  4. Find the information that tells which two quantities are equal.

  5. Use the information in (4) to write the equation.

  6. Solve the equation and check the solution to see that it satisfies the original problem.

  At this stage, the emphasis will be on translating statement problems into equations. Although some of the problems can be solved almost by inspection, the practice that we obtain in setting up equations will prove helpful in working more difficult problems.

Example 1.  If 2 times a certain integer is added to the next consecutive integer the result is 34. Find the integers.
Step 1. Reread!
Step 2. Let x be the first integer.
Step 3. Then x+ 1 is the next consecutive integer.
Step 4. 2 times a certain integer plus the next consecutive integer is 34.
Step 5. 2x+(x+1)=34

Step 6. Solve.  

  2x+(x+1)=34

  3x+1=34

  3x=33

  x=11

  Check.  2*11+(11+1)=34

Example 2.  Bob and Joe together earned $60. Both were paid at the same rate, but Bob worked three times as long as Joe. How much did each receive?

Step 1. Reread!

Step 2. Let x be the number of dollars that Joe received.

Step 3. Then 3x is the number of dollars that Bob received

Step 4. Bob and Joe together earned $60.
Step 5.3x+x=60

Step 6. Solve.  

  3x+x=60

  4x=60

  x=15

  3x=45

  Check  3*15+15=60

Example 3.  The sum of the digits of a two digit number is 12. If the digits are reversed the number is decreased by 36. What is the number?

Step 1. Reread!

Step 2. Let x be the tens digit.

Step 3. Then 12 - x is the units digit.

Step 4. If the digits are reversed then the number is decreased by 36

Step 5.10(12-x)+x = 10x+ (12-x) -36

Step 6. Solve.

  10(12-x)+x = 10x+ (12-x) -36

  =120-10x+x=10x+12-x-36

  =120-9x=9x-24

  =144=18x

  =x=8

  =12-x=4

  Therefore the number is 84.

  Check.  84-36=48

Example 4.  How many pounds of candy valued at 48¢ per pound should be added to 50 pounds of candy valued at 80¢ per pound in order for the store owner to be able to sell the candy at 60¢ per pound?
Step 1. Reread!
Step 2. Let x be the number of pounds of 48¢ per pound candy.
Step 3. Then 50+x would be the pounds of candy he would have at 60¢ per pound.
Step 4. The amount of candy at 48¢ per pound times 48¢, plus the amount of candy at 80¢ per pound times 80¢, must be equal to the amount of candy at 60¢ per pound times 60¢.
Step 5. (48¢/lb)(x lbs) + (80¢/lb) (50 lbs) = (60¢/lb) [(50+x)lbs]

Step 6. Solve.

  48x+80*50=60(50+x)

  48x+4000=3000+60x

  1000=12x

  x=(83(1)/3)lbs

  Check.  (83+1/3)48+80*50=60(50+83+1/3)

  Problems involving velocities (or speeds) will use the formula

  d=rt

  where d is the distance traveled, r is the rate, and t is the time. When the formula is used, d and r must be expressed in the same unit of distance, whiler and t must be expressed in the same unit of time.

Example 5.  A group of students drove to a lake in the north woods to fish. They traveled 380 miles in 7 hours, of which 4 hours were on a paved highway and the remaining time was on a dirt road. If the average speed on the dirt road was 25 miles per hour less than the average speed on the highway, then find for each part of the trip the average speed and the distance traveled.

Step 1. Reread!

Step 2. Let x be the speed on the dirt road.

Step 3. Then x+25 is the speed on the highway.

Step 4. The distance traveled on the highway plus the distance traveled on the dirt road is equal to 380 miles.

Step 5. Since d=rt, we have

  [(x+25)(mi)/(hr)](4hrs)+[x(mi)/(hr)](3hrs)=380mi

Step6. Solve.

  (x+25)4+3x=380

  4x+100+3x=380

  7x=280

  x=40 miles per hour

  x+25=65miles per hour 

  Check.  (40+25)4+40*3=380

  Work problems which involve the rate of performance can often be solved by first finding the fractional part of the task done by each person or machine in one unit of time, and then finding an equation that relates these various fractional parts.

Example 6.  A boy can cut a lawn in 4 hours while the father can cut it in 3 hours. How long would it take them to cut the same lawn working together?

Step 1. Reread!

Step 2. Let x be the number of hours that it would take them to cut the lawn Working together.

Step 3. Choose one hour as our unit of time. Now the boy can cut 1/4 of the lawn in one hour, the father can cut 1/3 of the lawn in one hour, and togeflner they can cut 1/x of the lawn in one hour.


Step 4. The amount cut by the boy in one hour plus the amount cut by the father in one hour is equal to the amount they can cut together in one hour.

Step 5.1/3+1/4=1/x

Step 6. Solve.

  1/3+1/4=1/x

  7/12=1/x

  x=12/7 hours

Sours: https://quickmath.com/math-tutorials/test-chapter-6.html
Linear equation word problem - Linear equations - Algebra I - Khan Academy

Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

yes y = 2x + 1
yes 5x = 6 + 3y
yes y/2 = 3 − x

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

line on a graph

The graph of y = 2x+1 is a straight line

 

  • When x increases, y increases twice as fast, so we need 2x
  • When x is 0, y is already 1. So +1 is also needed
  • And so: y = 2x + 1

Here are some example values:

xy = 2x + 1
-1y = 2 × (-1) + 1 = -1
0y = 2 × 0 + 1 = 1
1y = 2 × 1 + 1 = 3
2y = 2 × 2 + 1 = 5

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

Examples: These are NOT linear equations:

not y2 − 2 = 0
not 3√x − y = 6
not x3/2 = 16

Slope-Intercept Form

The most common form is the slope-intercept equation of a straight line:

y=mx+b graph

Example: y = 2x + 1

  • Slope: m = 2
  • Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

y − y1 = m(x − x1)

 Point-Slope Form

Example: y − 3 = (¼)(x − 2)

It is in the form y − y1 = m(x − x1) where:

  • y1 = 3
  • m = ¼
  • x1 = 2

General Form

And there is also the General Form of the equation of a straight line:

Ax + By + C = 0

(A and B cannot both be 0)

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function, with f(x) instead of y:

y = 2x − 3
f(x) = 2x − 3
These are the same!

And functions are not always written using f(x):

y = 2x − 3
w(u) = 2u − 3
h(z) = 2z − 3
These are also the same!

The Identity Function

There is a special linear function called the "Identity Function":

f(x) = x

And here is its graph:

Identity Function
It makes a 45° (its slope is 1)

It is called "Identity" because what comes out is identical to what goes in:

InOut
00
55
−2−2
...etc...etc

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

Constant Function

f(x) = C

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

 

Gradient (Slope) of a Straight LineY Intercept of a Straight LineDistance Between 2 PointsFinding Intercepts From an EquationGraph MenuAlgebra Menu

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Sours: https://www.mathsisfun.com/algebra/linear-equations.html

Equations help solving linear

Solving a System of Equations

Systems of linear equations take place when there is more than one related math expression. For example, in \(y = 3x + 7\), there is only one line with all the points on that line representing the solution set for the above equation.

When you are given 2 equations in the same question, and asked to solve for a unique answer, you can visualize the problem as be two lines on the same xy-plane. The following two equations are graphed on the same xy-plane:

$$ y = 3x + 5 $$ $$ y = - x $$

The solution set to any equation is the place where BOTH equations meet on the xy-plane. This meeting place is called the Point of Intersection. If you have a linear equation and a quadratic equation on the same xy-plane, there may be TWO POINTS where the graph of each equation will meet or intersect. Here's a geometric view:

Here is a sample of two equations with two unknown variables:

Example

$$ x + y = 10 $$ $$ 3x + 2y = 20 $$

There are three methods to solve our sample question.

  • 1) We can solve it graphically
  • 2) We can solve it algebraically
  • 3) We can also solve it through algebraic elimination

I will solve the question using all 3 methods. Method 1: Solve Graphically:

To solve graphically, it is best to write BOTH equations in the slope-intercept form or in the form: \(y = mx + b\) where m = the slope and b = the y-intercept as your first step. So, \(x + y = 10\) becomes \(y = - x + 10\) (slope-intercept form). Next, \(3x + 2y = 20\) becomes \(y = -\frac{3x}{2} + 10\) when written in slope-intercept form.

Then, graph the two lines, leading to the point of intersection. After graphing these lines, you'll find that BOTH equations meet at point (0,10). Point (0,10) means that if you plug x = 0 and y = 10 into BOTH original equations, you will find that it solves both equations. Here's what these two equations look like on the xy-plane:

Method 2: Solve algebraically

Steps:

1) Solve for eaither x or y in the first equation (\(x + y = 10\)). I will solve for y. So, \(x + y = 10\) becomes \(y = -x + 10\).

2) Plug the value of y (that is, -x + 10) in the second equation to find x. Our second equation was \(3x + 2y = 20\) and, after substituting, becomes \(3x + 2(-x + 10 ) = 20\)

Next: Solve for x.

$$ 3x -2x + 20 = 20 $$ $$ x + 20 = 20 $$ $$ x = 0 $$

3) Plug x = 0 into EITHER original equations to find the value of y. I will use our second equation.

$$ 3x + 2y = 20 $$ $$ 3(0) + 2y = 20 $$ $$ 0 + 2y = 20 $$ $$ y = 10 $$

So, our point of intersection is once again (0,10).

Method 3: Algebraic Elimination

This method deals with matching the variables to ELIMINATE or do away with one. Keep in mind that it is your choice which variable you want to eliminate first.

GOAL: Eliminate x and solve for y or vice-versa. Let's go back to our original equations.

In our second 3x + 2y = 20, you can eliminate 3x by multiplying -3 by EVERY term in our first equation (x + y = 10).

x + y = 10
3x + 2y = 20

-3(x) + -3(y) = -3(10)
3x + 2y = 20

-3x + -3y = -30
3x + 2y = 20

NOTICE that -3x and 3x are eliminated. See it? See why? Here's why: A negative PLUS a positive = ZERO.

We now have this:

-3y = -30
2y = 20

-3y + 2y = -30 + 20

-y = -10

y = 10.

Next: To find x, we plug y = 10 into EITHER of the original equations. By now you should see that our answer for x will be ZERO.

Here it is:

I will use x + y = 10

x + 10 = 10

x = 0.

Do you see what I see? Yes, I found the SAME point of intersection again, which is (0,10).

By Mr. Feliz
(c) 2005
Sours: https://www.freemathhelp.com/systems-linear-equations/
How To Solve Linear Equations In Algebra

Solving Equations: (lesson 1 of 4)

Solving Linear Equations

Equations with no parentheses

Example 1

Solve 5x - 4 - 2x + 3 = -7 - 3x + 5 + 2x

Solution 1

Step 1: Combine the similar terms

$5x - 4 - 2x + 3 = - 7 - 3x + 5 + 2x$
$3x - 1 = - x - 2$

Step 2: Add x to both sides.

$3x - 1 + x = - x - 2 + x$
$4x - 1 = - 2$

Step 3: Add 1 to both sides.

$4x - 1 + 1 = - 2 + 1$
$4x = - 1$

Step 4: Divide both sides by 4:

$\frac{{4x}}{4} = \frac{{ - 1}}{4}$
Solution is:$x = - \frac{1}{4}$

Exercise 1: Solve equations

Equations with parentheses

Example 2

Solve $2 (x - 4) + 4 (2 - x) = 5x - 4 (x + 1)$

Solution 2

Step 1: Simplify both sides:

$2 (x - 4) + 4 (2 - x) = 5x - 4 (x + 1)$
$2x - 8 + 8 - 4x = 5x - 4x - 4$
$- 2x = x - 4$

Step 2: Subtract x from both sides.

$- 2x - x = x - 4 - x$
$- 3x = - 4$

Step 3: Divide both sides by -3:

$\frac{{ - 3x}}{{ - 3}} = \frac{{ - 4}}{{ - 3}}$
Solution is:$x = \frac{4}{3}$

Exercise 2: Solve equations

Equations containing fractions

Example 3

Solve $x + \frac{1}{2} = \frac{x}{2} - \frac{2}{3}$

Solution 3

Step 1: Multiply both sides by the LCD. Lowest common multiple of 2 and 3 is 6. So, we multiply both sides by 6.

$6 \cdot \left( {x + \frac{1}{2}} \right) = 6 \cdot \left( {\frac{x}{2} - \frac{2}{3}} \right)$

Step 2: Remove brackets:

$6 \cdot x + 6 \cdot \frac{1}{2} = 6 \cdot \frac{x}{2} - 6 \cdot \frac{2}{3}$

Step3: This problem is similar to the previous

$6x + 3 = 3x - 4$
$6x + 3 - 3x = 3x - 4 - 3x$
$3x + 3 = - 4$
$3x + 3 - 3 = -4 - 3$
$3x = - 7$
$\frac{{3x}}{3} = \frac{{ - 7}}{3}$

Solution is:

$x = - \frac{7}{3}$

Exercise 3: Solve equations

More Complicated Example

Example 4

$\frac{{2x}}{{x + 2}} = \frac{4}{{x - 10}} + 2$

Solution 4

In this case the LCD is (x + 2)(x - 10). Here is the complete solution to this problem.

$$\frac{{2x}}{{x + 2}} = \frac{4}{{x - 10}} + 2$$ $$\cancel{{(x + 2)}}(x - 10)\frac{{2x}}{{\cancel{{x + 2}}}} = (x + 2)\cancel{{(x - 10)}}\frac{4}{{\cancel{{x - 10}}}} + (x + 2)(x - 10) \cdot 2$$ $$(x - 10) \cdot 2x = (x + 2) \cdot 4 + 2({x^2} + x - 10x - 20)$$ $$\cancel{{2{x^2}}} - 20x = 4x + 8 + \cancel{{2{x^2}}} + 2x - 20x - 40$$ $$ - 20x = 4x + 8 + 2x - 20x - 40$$ $$ - 20x = - 14x - 32$$ $$ - 20x + 14x = - 14x - 32 + 14x$$ $$ - 6x = - 32$$ $$x = \frac{{ - 32}}{{ - 6}}$$ $$x = \frac{{16}}{3}$$

Much more Complicated Example

Example 5

$\frac{1}{{x + 3}} = \frac{{ - 2x}}{{{x^2} + 5x + 6}}$

Solution 4

The first step is to factor the denominators

$\frac{1}{{x + 3}} = \frac{{ - 2x}}{{(x + 3)(x + 2)}}$

The LCD is (x + 3)(x + 2). The solution is:

$$\frac{1}{{x + 3}} = \frac{{ - 2x}}{{(x + 3)(x + 2)}}$$ $$(x + 3)(x + 2)\frac{1}{{x + 3}} = (x + 3)(x + 2)\frac{{ - 2x}}{{(x + 3)(x + 2)}}$$ $$x + 2 = - 2x$$ $$x + 2x = - 2$$ $$3x = - 2$$ $$x = - \frac{2}{3}$$

Exercise 4: Solve equations

Random Quote

Give me a place to stand, and I will move the earth.

Archimedes of Syracus

Random Quote

I like mathematics because it is not human and has nothing particular to do with this planet or with the whole accidental universe - because like Spinoza's God, it won't love us in return.

Bertrand Russell

Sours: https://www.mathportal.org/algebra/solving-equations/solving-linear-equations.php

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