Linear equations on the SAT test
SAT Subscore: Heart of Algebra
Studying linear equations
On the SAT test linear equations are part of heart of algebra subscore that includes 4 fundamental topics that appear in many SAT questions. The 4 topics are:
Linear functions– Start studying heart of algebra subscore with linear functions topic.
Linear equations– The current topic.
Systems of linear equations and linear inequalities– Continue to these last 2 topics after you study linear equations topic.
Linear equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.
Linear equations- summary
A linear equation is an algebraic equation in which each term has an exponent of one. It is called linear because it can be graphed as a straight line in the xy-plane. Since this is a fundamental topic it appears in many SAT questions.
Calculating an output of an expression requires calculating the value of an expression. In some questions we are given a word problem and in other questions we are given an equation that we must use it order to solve the expression.
Solving linear equations- In these questions we are given an equation and we are asked to solve it. Some questions may include fractions or absolute values.
Linear equations that don’t have one solution- In these questions we are asked to solve an equation that has no solution or an infinite number of solutions.
Creating and solving an equation from a word problem- In these questions we are given a word problem from which we need to write an equation and solve it.
The formula sheet for linear equations is listed below.
Linear equations formula sheet
Linear equations formula sheet is given below, it includes 3 types of basic formulas:
Calculations with negative numbers formulas– these are 5 very basic formulas that include multiplying and dividing positive and negative numbers.
Distributing and combining like terms formulas-one formula for opening parentheses (distributing) and one formula for creating them (combining).
Calculation with fraction formulas– includes three basic formulas for addition/ subtraction, multiplication and division of fractions.
Note that all the formulas are explained in detail on this page.
Calculating an output of an expression
In these questions we need to calculate the value of an expression. In some questions we are given a word problem and in other questions we are given an equation that we must use it order to solve the expression.
The skills required are:
Translating a word problem into a mathematical expression.
Solving an equation.
Performing calculations on an equation to match a given expression.
Before making any calculations, it is necessary to know the rules for calculations with negative numbers:
Negative * negative = positive. For example: -5*-4=20
Positive * negative = negative. For example: 5*-4=-20
Negative : negative = positive. For example: -8:-4=2
Positive : negative = negative. For example: 8:-4=-2
Negative : positive = negative. For example: -8:4=-2
Calculating an output of an expression given a word problem
Consider this example:
The length of the rectangle is 5 cm, the width of the rectangle is 20 percent smaller than its length.
What is the area of the rectangle?
We need to write an expression and calculate it. Rectangle’s area is the equals to its length multiplied by its width. If the length of a rectangle is 5 cm then the width of the rectangle is 5* 80%=4 cm. The area of the rectangle is 5*4=20 cm.
Calculating an expression given an equation
In these questions we are be asked to calculate an expression given an equation that we don’t need to solve. All we need in order to get the answer is to change the equation to match the expression.
Consider this example:
2x+10=3x+25
What is 6x+50 equal to?
Remember that if we need to divide or multiple, we need to do it for each component of the expression. In this case we can add parentheses to the expression. We see that the expression 6X+50 is 2 times larger than the expression 3x+25 therefore we can multiply the given equation by 2 and get the answer.
Before multiplying each side of the equation by 2 we add parentheses to the both sides of the equation:
(2x+10)*2=(3x+25)*2
4x+20=6x+50
So the answer is that 6x+50 is equal to 4x+20
Sometimes we will be asked to calculate an expression given an equation that we need to solve first.
Consider this example:
We know that 3x+10=25, calculate 4x+12.
Step 1: Solving the equation:
3x+10=25
3x=25-10
3x=15
x=5
Step 2: Calculating the expression:
4x+12=4*5+12=20+12=32
Solving a linear equation
In these questions we are given an equation and we are asked to solve it.
The skills required are solving an equation with one variable and making calculations including fractions, decimal fractions and absolute values.
In addition there are questions with two variables where we need to find the value of one variable from a word problem before solving for the second variable.
The rules for maintaining equality
Before starting to solve the equation, it is necessary to know the rules for maintaining equality:
1. We can add or subtract the same value from each side of the equation.
2. We can divide or multiply each side of the equation by a same value.
Consider this example:
2x+5=14
What is the solution to the given equation?
2x+5=14
2x+5-5=14-5
2x=9
2x:2=9:2
x=4.5
The steps for solving a linear equation
Distributing formula for removing parentheses:
(a+b)x=ax+bx
(a-b)x=ax-bx
For example: (2+x)5=10+5x
Combining like terms formula:
ax+bx=(a+b)x
ax-bx=(a-b)x
For example: 8x-2x=(8-2)x=6x
The goal of solving an equation is finding the x variable value. In order to do so we need to isolate x on the one side of the equation so that the other side of the equation will contain the constant (the answer).
Step 1: Removing parentheses and combining like terms in each side of the equation.
Distributing formula for removing parentheses: (a±b)x=ax±bx, for example (2+x)5=10+5x
Combining like terms formula: ax±bx=(a±b)x, for example 8x-2x=(8-2)x=6x
Step 2: Isolating the variable term on one side of the equation by using addition or subtraction.
Step 3: Solving the equation (finding x=) by using multiplication or division.
Step 4: Checking the answer by plugging the answer from the previous step into the equation.
Consider this example:
5x-10+x=2x+9(5+5)
What is the solution to the given equation?
Step 1: Removing parentheses and combining like term in each side of the equation
6x-10=2x+90
Step 2: Subtracting 2x and adding 10 to each side of the equation
6x-10-2x+10=2x+90-2x+10
4x=100
Step 3: Dividing each side of the equation by 4
4x:4=100:4
x=25
Step 4: Checking the answer
5*25-10+25=2*25+9(5+5)
125-10+25=50+9*10
140=140
Solving a linear equation with decimal fractions
Consider this example:
(2x-6)*0.5=(3x-10)*0.4-0.6
What is the solution to the given equation?
Solving the equation:
To solve the equation we need to open parentheses on both of its sides.
2*0.5x-6*0.5=3*0.4X-10*0.4-0.6
x-3=1.2x-4-0.6
-0.2x=-4-0.6+3
-0.2x=-1.6
x=1.6/0.2
x=8
Checking the answer:
(2*8-6)*0.5=10*0.5=5
(3*8-10)*0.4-0.6=14*0.4-0.6=5.6-0.6=5
Solving a linear equation with fractions
Addition and subtraction formula:
a⁄b ± c⁄b = (a±c)⁄b
For example: 1⁄4+51⁄4=51⁄2
Multiplication formula:
a⁄b* c⁄d= a*c⁄b*d
For example: 1⁄4*2=1*2⁄4*1=2⁄4 =1⁄2
Division formula:
a⁄b : c⁄d= a⁄b * d⁄c= a*d⁄b*c
For example: 1⁄4 : 2=1⁄4*1⁄2=1*1⁄4*2=1⁄8
To add or subtract fractions they must have the same denominator (the bottom value). We need to add or subtract the numerators and leave the denominator.
Addition and subtraction formula: a⁄b±c⁄b=(a±c)⁄b. For example: 1⁄4+51⁄4=51⁄2
To multiply fractions, you have to multiply the nominators and multiply the denominators.
Multiplication formula: a⁄b*c⁄d=a*c⁄b*d. For example: 1⁄4*2=1*2⁄4*1=2⁄4 =1⁄2
To divide fractions, you must flip the second fraction and then multiply it with the first fraction.
Division formula: a⁄b : c⁄d=a⁄b*d⁄c=a*d⁄b*c. For example: 1⁄4 : 2=1⁄4*1⁄2=1*1⁄4*2=1⁄8
If the denominators are different we can first reach a common denominator in order to cancel the denominators and then we can continue to solve the equation. This is done by multiplying by the smallest common multiple of the denominators of the fractions.
Consider this example:
31⁄2x-21⁄4=2x+51⁄4
What is the solution to the given equation?
Solving the equation:
The common denominator for the numbers 2 and 4 is 4 so we need to multiply the equation by 4.
4*(31⁄2x-21⁄4)=4*(2x+51⁄4)
4*3x +4*1⁄2x-4*2-4*1⁄4=8x+4*5+ 4*1⁄4
12x+2x-8-1=8x+20+1
14x-9=8x+21
6x=30
x=5
Checking the solution:
31⁄2x-21⁄4=2x+51⁄4
31⁄2*5-21⁄4=2*5+51⁄4
15+2+1⁄2-21⁄4=151⁄4
17+1⁄2-21⁄4=151⁄4
151⁄4=151⁄4
Solving equations with absolute values
Pay attention that absolute value calculations are rare on the SAT.
The absolute value of a negative number is always positive. Therefore, we need to split an absolute value equation into 2 equations.
Consider this example:
What is the solution for the equation |6x-10|=2?
Solving the equation:
Since we don’t know the sign of |6x-10|=2, we need to write 2 equations.
Equation 1: (6x-10)=2
6x=12
x=2
Equation 2: (6x-10)=-2
6x=8
x=8⁄6=12⁄6=11⁄3
Checking the answers:
|6*2-10|=2
2=2
|6*8⁄6-10|=2
|48⁄6-10|=2
|8-10|=2
|-2|=2
Solving an equation with two variables given an input of one variable
In these questions we are given a word problem and an equation with two variables. We need to find the value of one variable inside the word problem and then plug it into the equation. After that we will have an equation with one variable that we can solve.
Consider this example:
0.6x+0.4y=86
The equation above represents a weighted average of 86 in two subjects, x is defined as a grade in course 1 and y is defined as a grade in course 2.
If the first grade was 90, what was the second grade?
The first grade should be placed in the equation instead of x and then we will be able to solve an equation with 1 variable y.
Plugging the input into the equation:
0.6*90+0.4y=86
Solving the equation with 1 variable:
54+0.4y=86
0.4y=32
y=80
The second grade was 80.
Checking the answer:
In order to test the answer we will place it in the equation: 0.6*90+0.4*80=54+32=86.
Linear equations that don't have one solution SAT questions
Previous examples included common equations that had one single solution.
In these questions you are asked to solve an equation that has no solution or an infinite number of solutions.
The skills required are solving an equation with one variable and identifying the conditions for no solution or an infinite number of solutions to the equation.
Linear equations with no solution
Equations with no solution appear when the x variable is eliminated from the equation so what is left is only 2 constants a=b. Since a is a different from b this is a false statement and the equation has no solution.
Consider this example:
What is the solution for the equation 6x+12=3(2x+2)?
6x+12=3(2x+2)
6x+12=6x+6
12-6=6x-6x
6=0 is a false statement therefore the equation has no solution.
Finding the condition for no solution:
In this type of questions we are given an equation with a as a constant and asked for which value of a there is no solution to the equation. In order to find the solution, we need to solve the equation. Another possibility is to place the answers into the equation and see if we get a false statement.
Consider this example:
For which value of a the equation 6ax+12=3(2x+2) has no solution? a=0,1,2?
Solving the equation:
6ax+12=3(2x+2)
6ax+12=6x+6
6ax-6x=-6
6x(a-1)=-6
x(a-1)=-1
x=-1/(a-1)
a≠1 since division by zero is not allowed therefore there is no solution when a=1.
Placing given answers into the equation:
For a=0 we will get
12=6x+6
6x=6
x=1
We can check this answer by plugging x=1 and a=0 into the equation to see if the quality holds:
6*0*1+12=3(2*1+2)
0+12=3*4
12=12
For a=1 we will get
6x+12=6x+4
12=4 is a false statement therefore there is no solution when a=1.
For a=2 we will get
12x+12=6x+6
6x=-6
x=-1
We can check this answer by plugging x=-1 and a=2 into the equation to see if the quality holds:
6*2*-1+12=3(2*-1+2)
-12+12=3*(-2+2)
0=3*0
0=0
Linear equations with an infinite number of solutions
Equation with an infinite number of solutions appears when the constants are eliminated from the equation so what is left is only 2 x variables in the form x=x. This is a true statement for every x therefore the equation has an infinite number of solutions.
Consider this example:
What is the solution for the equation 6x+12=3(2x+4)?
6x+12=3(2x+4)
6x+12=6x+12
6x=6x
x=x
x=x is a true statement for every x therefore the equation has an infinite number of solutions.
We can also continue to solve:
x=x
x-x=0
0=0 is also a true statement for every x
Creating and solving an equation from a word problem
In these questions we are given a word problem from which we need to write an equation and solve it.
The skills required are defining the x variable in a word problem, writing an equation from a word problem and solving an equation with one variable.
Consider this example:
The inventory of a store contained bags with balls received from the supplier so that in each bag were packed 10 balls. The store worker decided to pack the balls in smaller packages in order to sell them to customers. He took a certain number of bags and emptied the balls from them. He then added to the balls he took out another 30 balls left in the store from previous orders and then he divided these balls into 75 bags so that each bag contained 2 balls.
The worker noticed that he forgot to write down how many bags he had taken from stock. What was the amount?
In this example we will need to write an equation in order to reach a solution.
Step 1: X definition: Let’s look at the question and mark in x the number of bags the employee took from stock.
The worker took x bags containing 10 balls in each bag and added them 30 balls: 10x+30
Then he divided these the balls into 75 bags so that each bag contained 2 balls: 75*2
Step 2: writing the expressions in each side of equation: We need to find 2 equal expressions.
Since we know that the worker sorted the same amount of balls, we know what to write on each side of the equation:
10x+30=75*2.
Step 3: Solving the equation:
10x+30=75*2
10x=150-30
10x=120
x=12
The solution is x=12.
Solving a word problem using two equations
What if the question is not about is the x variable value? In this case after calculating x value, we will write another equation using y variable (y will be defined as what is asked in the question).
The steps for writing a second equation:
Step 1: Looking at the relevant data parts again, this time ignoring the data we needed to calculate x.
Step 2: Identifying parts of the data which will give us an equality.
Step 1+2: Writing the second equation consisting y variable and solving it.
Let’s add to the above example: There were 50 bags with balls in the store’s inventory.
I addition, let’s assume that the question is not to find the amount off bags that the worker took from the store’s inventory. Instead, the question is how many bags were left in the inventory?
The inventory of a store contained bags with balls received from the supplier so that in each bag were packed 10 balls. There were 50 bags with balls in the store’s inventory. The store worker decided to pack the balls in smaller packages in order to sell them to customers. He took a certain number of bags (we solved x=12) and emptied the balls from them. He then added to the balls he took out another 30 balls left in the store from previous orders and then he divided these balls into 75 bags so that each bag contained 2 balls. How many bags were left in the inventory?
The first equation was 10x+30=75*2. We solved it so x=12.
Writing a second equation: We will use the letter y for the new variable. The y variable will be according to the question: y=the number off bags that were in the inventory. The equation will be: y=50-x. We had already solved x=12, so now we know that y=38.
The question could also be even more complicated: How many bags were left in the inventory? In this case the second equation should be y=(50-12)*10 and the answer will be 380 balls.
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