Linear Equation Flashcards
1
Q
a mathematical statement that tells you that two quantities are equal in value.
A
Equation
2
Q
Example:Which of the following are equations?
a. 2x + 3 = – 9
b. x = – 7
c. x – 5
A
Solution:The mathematical statements inaandbare equations because they have an equal sign. On the other hand,cis not an equation because of the absence of an equal sign
3
Q
quantities on the left of the equal sign.
A
The left-hand side of the equation
4
Q
quantities on the right of the equal sign.
A
The right-hand side of the equation
5
Q
Example:Is x = 5 the solution to x + 2 = 7?
A
Solution:Yes, because if we substitute x = 5 to x + 2 = 7:
x + 2 = 7
(5) + 2 = 7
7 = 7
The left-hand side and the right-hand side of the equation are equal. Indeed, x = 5 is the solution to x + 2 = 7.
6
Q
rules or principles that allow us to manipulate equations so we can determine the values of the unknown variable.
A
properties of equality
7
Q
This property is pretty obvious and logical. The value of a number is always equal to itself.
For instance, 1020 will always be equal to 1020. If someone tells you that 1020 = 1100, he is logically false since 1020 is always equal to 1020
A
Reflexive Property of Equality
8
Q
This property tells us that in an equation if we switch the positions of the quantities on the left-hand side and the right-hand side of the equation, the equation will still hold. This also implies that both sides of the equation are of the same value.
A
Symmetric Property of Equality
9
Q
tells us that if a quantity is equal to a second quantity, and if the second quantity is equal to a third quantity, then we can conclude that the first quantity is equal to the third quantity.
A
Transitive Property of Equality
For any real numbers p, q, and r:
If p = q and q = r, then p = r
10
Q
tells us that if we add a certain number to two equal quantities, the result will still be equal.
A
Addition Property of Equality (APE)
11
Q
tells us that if we subtract two equal quantities by the same number, the results will still be equal.
A
Subtraction Property of Equality (SPE)
12
Q
tells us that the results will still be equal if we multiply two equal quantities by the same number.
For example, we know that 2 + 2 = 3 + 1. Suppose that we multiply both sides of this equation by 5:
As shown above, the results will still be equal even after multiplying both sides by the same number.
A
Multiplication Property of Equality
13
Q
This property tells us that the results will still be equal if we divide two equal quantities by the same number (that number can be any number but must not be equal to 0).
A
Division Property of Equality
14
Q
This property tells us that multiplying the sum of two or more addends is equal to the result when we multiply the addends by that number and add them.
A
Distributive Property of Equality
15
Q
If x = y, then either x or y can be substituted into any equation for the other.
A
Substitution Property of Equality
16
Q
p = p
A
Reflexive Property
17
Q
If p = q, then q = p
A
Symmetric Property
18
Q
If p = q and q = r, then p = r
A
Transitive Property
19
Q
If p = q, then p + r = q + r
A
Addition Property of Equality
20
Q
If p = q, then p – r = q – r
A
Subtraction Property of Equality
21
Q
If p = q, then pr = qr
A
Multiplication Property of Equality
22
Q
If p = q, then p/r = q/r where r0
A
Division Property of Equality
23
Q
p(q + r) = pq + pr
A
Distributive Property of Equality
24
Q
If x = y and ax + by = c, then ax + bx = c or ay + by = c
A
Substitution Property of Equality
25
Example: Which of the following are linear equations in one variable?
a. 2x + 7 = 19
b. x^2 + 6x + 9 = 0
c. x + y = 2
Solution: The equation in a is the only linear equation in one variable among the given equations since the highest exponent of its variable is 1 and it has only one variable. The equation in b is not a linear equation since the highest exponent of its variable is 2 (it is a quadratic equation). Meanwhile, the equation in c, although the highest exponent of its variable is 1, is not a linear equation in one variable because there are two variables involved (i.e., x and y).
26
Example 1: Let us try to solve for the value of x in x – 9 = 10.
Solution: To find the value of x, our goal is to isolate the variable from the constants. This means that if we want to solve for x, then x must be the only quantity on the left side of the equation and the other quantities must be on the right side. But how can we achieve that?
x will be the only quantity on the left if we get rid of -9 on the left side. How can then we remove -9 on the left side?
The addition property of equality (APE) states that we can add the same number to both sides of the equation.
Applying the APE, we can add 9 to both sides of the equation so we can cancel -9 on the left side:
Now, it is seen that x = 19.
That’s it! We have solved the value of x in x – 9 = 10. The answer is x = 19.
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Example 2: Solve for x in x – 12 = 22
Solution:
x – 12 = 22
x – 12 + 12 = 22 + 12 (adding 12 to both sides of the equation)
x = 34
Hence, x = 34.
Note that the explanation for why it is valid to add 12 to both sides of the equation is that we apply the addition property of equality.
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Example 3: Solve for x in x + 10 = 52
Solution: To isolate x from the constants, we must get rid of 10 by subtracting 10 from both sides of the equation. Subtracting the same number from both sides of the equation is valid because of the subtraction property of equality (SPE) discussed earlier.
x + 10 = 52
x + 10 – 10 = 52 – 10 (subtracting 10 from both sides of the equation)
x = 42
Thus, the answer is x = 42
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To isolate x from the constants, we can transpose the constant to the right-hand side of the equation so that x will be the only quantity that will remain on the left side.
Transposition Method
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Example 1: Let us solve x + 9 = 10 using the transposition method.
Solution:
Our goal is to isolate x from other constants by transposing 9 to the right-hand side of the equation. Once a quantity “crosses” the equality sign, its sign reverses (i.e., from positive 9 to -9).
After we transpose 9 to the right-hand side and reverse its sign, we add it to the quantity on the right-hand side (which is 10).
Then, we perform some arithmetic:
Thus, the answer is x = 1.
31
Example 2: Use the transposition method to solve for x in x – 9 = 12.
Solution: Transposing -9 to the right-hand side will reverse its sign (i.e., from negative to positive):
x = 9 + 12
x = 21
Thus, the answer is x = 21.
32
Example 3: Solve for x in x + 6 = 5 using the transposition method.
Solution: Transposing 6 to the right-hand side will reverse its sign (i.e., from positive to negative):
x = – 6 + 5
x = – 1
Thus, the answer is x = -1.
33
Example 4: Solve for x in x – 4 = – 9 using the transposition method.
Solution:
x – 4 = -9
x = 4 + (- 9) (transposing -4 to the right-hand side will change its sign to positive)
x = -5
Thus, the answer is x = -5.
Note: In this review, we will be using the transposition method more frequently to isolate x from other quantities. The transposition method is more convenient than adding numbers to or subtracting numbers from both sides of the equation.
34
Example 1: Let us try to solve for x in 2x + 4 = 6.
Solution: Again, to solve for x in an equation, it must be isolated from the constants or x should be the only quantity on the left-hand side of the equation.
Let us start by getting rid of 4 on the left-hand side by using the transposition method:
What is left is 2x = 2. Again, our goal is to make x the only quantity on the left-hand side. This means that we need to cancel out 2 in 2x. But how do we cancel it?
We can divide both sides of the equation by 2 so that 2 will be canceled in 2x. This is valid because the division property of equality guarantees us that dividing both sides of the equation by the same number will preserve equality.
As we can see, the answer is x = 1.
Here’s a quick preview of what we have done above:
2x + 4 = 6
2x = -4 + 6 (transposing 4 to the right-hand side will turn it into -4)
2x = 2
2x⁄2 = 2⁄2 (dividing both sides of the equation by 2)
x = 1
Thus, the solution to 2x + 4 = 6 is x = 1
You can verify that x = 1 is the solution by substituting it back to 2x + 4 = 6. Notice that the equation will be true if x = 1:
2(1) + 4 = 6
2 + 4 = 6
6 = 6
35
Example 2: Solve for x in 3x – 18 = 27
Solution: To solve for x, x should be the only quantity on the left-hand side.
We start by transposing -18 to the right-hand side. If we transpose it, it will have a positive sign.
3x = 18 + 27
3x = 45
To cancel out 3 in 3x, we divide both sides of the equation by 3:
3x⁄3= 45⁄3
x = 15
Therefore, the answer is x = 15
36
Example 3: Solve for x in 4x – 18 = 2
Solution: To solve for x, x should be the only quantity on the left-hand side.
We start by transposing -18 to the right-hand side. If we transpose it, it will change its sign from negative to positive.
4x = 18 + 2
4x = 20
To cancel out 4 in 4x, we divide both sides of the equation by 4:
4x⁄4= 20⁄4
x = 5
Therefore, the answer is x = 5.
37
Example 4: Solve for x in 7x + 2 = 16
7x + 2 = 16
7x = -2 + 16 Transposition Method (we transpose 2 to the right-hand side)
7x⁄7= 14⁄7 Division Property of Equality (divide both sides of the equation by 7)
x = 2
38
Example 1: Solve for x in 3x – 3 = x + 5
Solution: Let us put all x first on the left-hand side. We can do this by transposing the x on the right-hand side to the left-hand side. Like numbers, variables will also reverse their sign once they cross the equality sign.
We can then combine 3x and -x to obtain 2x:
Now, we have 2x – 3 = 5. We can apply the techniques we have learned above to solve this one:
2x – 3 = 5
2x = 3 + 5 Transposition Method
2x = 8
2x⁄2= 8⁄2 Division Property of Equality
x = 4
39
Example 2: Solve for 9 – x = 2x – 3
Solution: We start by putting all x on the left-hand side of the equation using the transposition method.
We can then combine -2x and -x to obtain -3x:
Thus, we have –3x + 9 = – 3. Let us now use the techniques we have learned to solve for x:
-3x + 9 = -3
-3x = -9 + (-3) Transposition Method
-3x = -12
–3x⁄-3= 12⁄-3 Division Property of Equality
x = – 4
40
Example 3: Solve for 3(2x + 1) = 15
Solution: Since 3 is multiplied by the sum of addends, we can apply the distributive property so that our equation will be in the form ax + b = c.
3(2x + 1) = 15
3(2x) + 3(1) = 15 Distribute 3 to 3(2x + 1)
6x + 3 = 15
Now, let us continue the process using the techniques we have learned in the previous sections:
6x = -3 + 15 Transposition Method
6x = 12
6x⁄6= 12⁄6 Division Property of Equality
x = 2
Therefore, the answer is x = 2
41
Example 4: Solve for x in
3x + 2/2=1/3
Solution: In case a linear equation in one variable is fractional in form, we “remove” the denominator by multiplying both sides of the equation by the Least Common Denominator (this method is valid because of the multiplication property of equality).
The Least Common Denominator (LCD) is the lowest common multiple of the denominators 3 and 2. Therefore, the LCD should be 6.
We then multiply both sides of the equation by the LCD (which is 6):
Now, our equation becomes 3(3x + 2) = 2
Let us continue solving for x:
x = -4/9
Therefore, the answer is x = -4/9
42
Example 5: Solve for x in
x+4/2=1/4
Solution:
x = -7⁄2
43
To solve word problems using linear equations, follow these steps:
Read and understand the given problem and determine what is being asked.Represent the unknown in the problem using a variable.Construct a linear equation that will describe the problem.Solve for the value of the unknown variable in the linear equation.
44
Example 1: The sum of a number and 5 is – 3. What is the number?
Solution:
Step 1: Read and understand the given problem and determine what is being asked. The problem is asking us to determine the number such that the sum of that number and 5 is – 3.
Step 2: Represent the unknown in the problem using a variable. Let x represent the number we are looking for.
Step 3: Construct a linear equation that will describe the problem. The problem states that the sum of the unknown number (represented by x) and 5 is – 3. Therefore, we construct the linear equation below:
x + 5 = – 3
Step 4: Solve for the value of the unknown variable in the linear equation. Using the equation we have derived from Step 3, we solve for the value of x:
x + 5 = – 3
x = – 5 + (-3) Transposition Method
x = -8
Thus, the number is -8.
45
Example 2: Fred has 52 books in his collection. He gave some of these books to Claude. Fred also gave some books to Franz. The number of books that Fred gave to Franz is twice the number of books that he gave to Claude. The number of books left to Fred after he gave some to Claude and Franz is 22. How many books did Claude receive?
Solution:
Step 1: Read and understand the given problem and determine what is being asked. The problem is asking us to determine the number of books Claude received from Fred.
Step 2: Represent the unknown in the problem using a variable. Let x be the number of books that Claude received. Since Franz received twice the number of books Claude received, we let 2x be the number of books Franz received.
To summarize:
x = number of books that Claude received2x = number of books that Franz received
Step 3: Construct a linear equation that will describe the problem. It’s stated that after Fred gave some books to Claude and Franz, there were only 22 books left.
We can express this statement this way:
52 – (number of books that Claude received) – (number of books that Franz received) = 22
Using the variables we have set in Step 2:
52 – x – 2x = 22
Step 4: Solve for the value of the unknown variable in the linear equation.
52 – x – 2x = 22
52 – 3x = 22 Combining like terms
-3x = -52 + 22 Transposition Method
-3x = -30
-3x⁄-3= -30⁄-3 Division Property of Equality
x = 10
Since x represents the number of books Claude received from Fred, then Claude received 10 books from Fred.
Using the value of x that we have obtained in the problem, can you determine how many books Franz received from Fred?
Yes, the answer is 20 since Franz received twice the number of books Claude received.
46
Example 3: The total number of participants in a mini-concert by a local band is 300. The number of female participants in the mini-concert is half the number of male participants in the event. How many male participants are there in the mini-concert?
Solution:
Step 1: Read and understand the given problem and determine what is being asked. The problem is asking us to determine the number of male participants in the mini-concert.
Step 2: Represent the unknown in the problem using a variable. Let x be the number of male participants in the mini-concert. Since the number of female participants in the mini-concert is half the number of male participants, we let ½ x represent the number of female participants in the event.
Step 3: Construct a linear equation that will describe the problem. The total number of participants in the mini-concert is 300. We can express this as:
(Number of Male Participants) + (Number of Female Participants) = 300
Using the variables we have set in Step 2:
x + ½ x = 300
Step 4: Solve for the value of the unknown variable in the linear equation.
Let us solve for x in x + ½ x = 300
x + ½ x = 300
2(x + ½ x) = 2(300) Multiplying both sides of the equation by the LCD
2(x) + 2(½ x) = 600 Distributive Property
2x + x = 600
3x = 600
3x⁄3= 600⁄3 Division Property of Equality
x = 200
Since x represents the number of male participants in the mini-concert, there are 200 male participants.
47
composed of two or more linear equations. The solution of a system of linear equations will satisfy all of the equations in the system.
system of linear equations
48
To solve a system of linear equations using the substitution method, follow these steps:
Solve for the value of one variable in one of the linear equations in terms of the other variable. Substitute the expression for the variable you have obtained in Step 1 in the other linear equation.Solve for the value of the other variable in the equation you have obtained from Step 2.Plug in the value of the unknown variable you have computed in Step 3 in the expression you have obtained in Step 1 to find the value of the other variable.
49
Example 1: Solve for the values of x and y that will satisfy x + y = 9 and x – y = 3
Let us write first the given equations:
Equation 1: x + y = 9
Equation 2: x – y = 3
Step 1: Solve the value of one variable in one of the linear equations in terms of the other variable. Using Equation 1, we solve for the value of y in terms of x. This means we let y be the only quantity on the left-hand side while the other quantities must be on the right-hand side, including x. To make this possible, we just transpose x to the right side:
x + y = 9 ⟶ y = – x + 9
Step 2: Substitute the expression for the variable you have obtained in Step 1 in the other linear equation. We have obtained y = -x + 9 in Step 1. What we are going to do is to substitute this value of y into the y in Equation 2:
x – y = 3 (Equation 2)
x – (-x + 9) = 3 (We substitute y = -x + 9)
Notice that once we substitute y = -x + 9 in Equation 2, Equation 2 will now be a linear equation in one variable.
Step 3: Solve for the value of the other variable in the equation you have obtained from Step 2. The equation we have obtained in Step 2 is x – (-x + 9) = 3. Our goal now is to solve for x.
We just use the techniques in solving linear equations in one variable:
x – (-x + 9) = 3
x + x – 9 = 3 Distributive Property
2x – 9 = 3
2x = 9 + 3 Transposition Method
2x = 12
2x⁄2= 12⁄2 Division Property of Equality
x = 6
Now that we have obtained the value for x which is x = 6, let us solve for y.
Step 4: Plug in the value of the unknown variable you have computed in Step 3 in the expression you have obtained in Step 1 to find the value of the other variable. From Step 3, we have obtained x = 6. We substitute x to the equation we obtained in Step 1, y = -x + 9.
y = -x + 9 (The expression we have obtained in Step 1)
y = -(6) + 9 (Substitute x = 6 which we have obtained in Step 3)
y = 3
That’s it! The solution for our system of linear equations is x = 6 and y = 3.
50
To solve a system of linear equations using the elimination method, follow these steps:
Write the given equations in standard form.Add or subtract the given equations so that one variable will be eliminated. If there’s no variable that can be eliminated by adding or subtracting the equations, you may multiply an equation by a constant to allow the elimination of a variable.Solve for the value of the remaining variable.Substitute the value of the variable you have computed in Step 3 to any of the given equations then solve for the value of the other variable.
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Example 1: Solve for the values of x and y that will satisfy x + y = 10 and x – y = 12
Solution:
Step 1: Write the given equations in standard form. If you can recall, the standard form of a linear equation in two variables is ax + by = c. Both x + y = 10 and x – y = 12 are already in standard form, so we can skip this step.
Step 2: Add or subtract the given equations so that one variable will be eliminated. If there’s no variable that can be eliminated by adding or subtracting the equations, you may multiply an equation by a constant to allow the elimination of a variable.
If we add the equations x + y = 10 and x – y = 12, the y variable will be eliminated. There’s no need to multiply the equations with a constant since we can immediately cancel a variable just by adding the equations.
After adding the equations, the resulting equation will be 2x = 22
Step 3: Solve for the value of the remaining variable. The remaining variable in 2x = 22 is x. We solve for x in this step by dividing both sides of 2x = 22 by 2:
Thus, x = 11
Step 4: Substitute the value of the variable you have computed in Step 3 to any of the given equations then solve for the value of the other variable. We substitute x = 11 to one of the given equations. Let us use x + y = 10:
x + y = 10
(11) + y = 10 Substituting x = 11
y = -11 + 10 Transposition Method
y = -1
Therefore, the solution for the system of linear equations is x = 11 and y = -1.
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1) What must be the value of x to satisfy x - 9 = 21?
a) 30
b) 31
c) 32
d) 33
A
53
2) Solve for x: 2(x - 3) = - 4
a) 0
b) 1
c) 2
d) 3
B
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3) A number is increased by 2 and the result is equal to twice of that number. What is the
number?
a) 1
b) 0
c) 3
d) 2
D
55
4) Solve for the value of x: 3x - 10 = 17
a) 9
b) 12
c) 15
d) 19
A
56
5) Solve for the value of x in 2 + 𝑥/9 = 1
a) 6
b) 5
c) 7
d) 11
C
