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Algebraic expressionSimplifying expressions
Expanding expression Evaluate the expression ExponentsHere you will learn about math equations, including what they are and how to solve them.
Students will first learn about math equations as part of expressions and equations in 6th grade. They continue to expand upon this knowledge in 7th and 8th grade.
Math equations are statements made up of algebraic expressions and an equal sign. Math equations include defined math formulas that can be solved with substitution.
For example, here is a rectangle with base b and height h.
The area of any rectangle can be found with the formula b \times h.
Step-by-step guide: Math formulas
For example,
Does x = 2 make the equation 5x+2=14 true?
Substitute 2 for x and follow the order of operations:
5 \times 2+2=14
10+2=14
12
Substituting 2 for x, does not make the equation true, so 2 is not a solution of the equation 5x+2=14.
Step-by-step guide: Substitution
One step equations are algebraic equations with a single variable that can be solved with one step.
In order to solve, you will need to isolate the variable (get it alone) on one side of the equation. This can be done with a model or by using the inverse operation.
For example, 3x =15
Using a model: | Using inverse operations: |
---|---|
Sort the (+1) tiles so an equal amount are with each (x) tile. There are five (+1) tiles for each (x) tile. x=5 | 3x=15 Get the variable alone by using the inverse operation. The inverse operation of multiplication is division. Divide 3 on both sides of the equation. \cfrac{3x}{3}=\cfrac{15}{3} Since \cfrac{3}{3} \,=1, the left side can be written as 1x or x. The right side of the equation can be written as \cfrac{15}{3}=5. x=5 You can check your answer by substituting the value for x into the original equation. \begin{aligned} 3x&=15\\ 3(5)&=15 \; \color{#77dd77} ✔ \end{aligned} |
Step-by-step guide: One step equations
To solve equations with fractions, apply the inverse operation to both sides of the equation – a strategy also referred to as the “balance method.”
The inverse operation of division is multiplication.
For example, \, \cfrac{x+3}{3}=8
3\left(\cfrac{x+3}{3}\right)=3(8) \quad *Multiply each side by 3.
x+3=24
-3 \quad \; -3 \quad *Subtract 3 from both sides.
x = 21
Check your work by substituting the answer into the equation.
Step-by-step guide: Solve equations with fractions
How does this relate to 6th grade math, 7th grade math, and 8th grade math?
There are a lot of ways to use math equations. For more specific step-by-step guides, check out the pages linked in the “What are math equations?” section above or read through the examples below.
Assess math progress for the end of grade 4 and grade 5 or prepare for state assessments with these mixed topic, multiple choice questions and extended response questions!
DOWNLOAD FREEAssess math progress for the end of grade 4 and grade 5 or prepare for state assessments with these mixed topic, multiple choice questions and extended response questions!
DOWNLOAD FREEFind the volume of the cube using the formula, V=s \times s \times s or V=s^3
Use the formula V=s \times s \times s to find the volume of the cube.
The sides of the cube are all 8. Substitute the measurements into the given formula.
V=8 \times 8 \times 8
2Work carefully to answer the question, one step at a time.
\begin{aligned} & V=8 \times 8 \times 8 \\\\ & V=64 \times 8 \\\\ & V=512 \end{aligned}
3Write the final answer clearly.
The final answer is 512 units^3.
Does r=-2 make the inequality r \leq-11 true?
Substitute each variable with its given value.
-2 \leq-11
Solve using the order of operations.
There is nothing to calculate, but you read the inequality ‘-2 is less than or equal to -11.’ Also consider the inequality on a number line.
Since -2 is not less than or equal to -11, substituting -2 for r makes the inequality false. This means -2 is NOT a solution for r \leq-11.
*Note that there are an infinite amount of solutions for this inequality – any number less than or equal to -11 makes the inequality true.
Find the value of e \, (114-t) when e=11 and t=58.
Substitute each variable with its given value.
11 \, (114-58)
Solve using the order of operations.
11 \, (114-58)
=11 \, (56) \quad \quad **Remember 11 \, (56) is the same as 11 \times 56.
=616
Solve the equation for x.
8+x=-13
Identify the inverse operation to use.
8+x=-13
Because of the commutative property, this equation could also be written as:
x+8=-13 → inverse operation is subtraction because subtraction undoes addition.
Do the inverse operation to both sides of the equation.
8+x =-13
-8 \quad\; -8
Solve for the unknown variable.
On the left side of the equation 8-8=0, leaving just 0+ x, which is the same as x.
On the right side of the equation:
\begin{aligned}
x= & -13 \\\\
& -8 \\\\
x= & -21
\end{aligned}
Check the answer.
Solve the equation, \, \cfrac{x}{5}=-20 for x.
Identify the inverse operation to use.
\cfrac{x}{5}=-20 → inverse operation of division is multiplication.
Do the inverse operation on both sides of the equation.
\cfrac{x}{5} \times 5=-20 \times 5
Solve for the unknown variable.
On the left side of the equation: \cfrac{x}{5} \times 5=\cfrac{5 x}{5}=1 x
On the right side of the equation: -20 \times 5=-100
\begin{aligned} \cfrac{x}{5} \times 5 & =-20 \times 5 \\\\ 1 x & =-100 \\\\ x & =-100 \end{aligned}
Check the answer.
Solve for x\text{: } \cfrac{x-9}{8}=6.5
Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, 9 is subtracted from x and then divided by 8 (the denominator of the fraction).
\cfrac{x-9}{8}
Apply the inverse operations, one at a time, to both sides of the equation.
First, multiplying both sides of the equation by 8.
Then, add 9 to each side.
8\left(\cfrac{x-9}{8}\right)=(6.5) \, 8 \quad *Multiply each side by 8.
x-9=52
+9 \quad \quad +9 \quad *Add 9 to both sides.
x = 61
Write the final answer, checking that it is correct.
The final answer is x=61.
You can check the answer by substituting the answer back into the original equation.
\cfrac{x-9}{8}=\cfrac{61-9}{8}=52 \div 8=6.5
1. Find the area of the triangle using the appropriate formula.
Use the formula A=\cfrac{1}{2} \, b h to find the area of the triangle.
Substitute the measurements into the formula and solve.
\begin{aligned} & A=\cfrac{1}{2} \, (4 \times 6) \\\\ & A=\cfrac{1}{2} \, (24) \\\\ & A=12 \end{aligned}
The area of the triangle is 12 \, m^2.
2. Find the value of 45+88x when x=13.
Substitute each variable with its given value.
45+88 \times 13
Solve using the order of operations.
\begin{aligned} & 45+88 \times 13 \\\\ & =45+1,144 \\\\ & =1,189 \end{aligned}
3. Which number is a solution for t > -56 \, ?
Substitute each variable with its given value and then read each inequality:
-55 > -56… ‘-55 is greater than -56’
-56 > -56… ‘-56 is greater than -56’
-57 > -56… ‘-57 is greater than -56’
-258 > -56… ‘-258 is greater than -56’
Substituting -55 for t makes the inequality true, so -55 is a solution for t > -56.
4. Solve the equation for x \text{: } -4x=22
-4x=22 → The inverse operation is division, so divide both sides of the equation by -4.
\cfrac{-4 x}{-4}=\cfrac{22}{-4}
On the left side of the equation: \cfrac{-4 x}{-4}=1 x and 1 x=x
On the right side of the equation: \cfrac{22}{-4}=-5.5
\begin{aligned} & \cfrac{-4 x}{-4}=\cfrac{22}{-4} \\\\ & x=-5.5 \end{aligned}
Check:
5. Solve the equation for p \text{: } p-\cfrac{4}{5}=\cfrac{1}{3}
p-\cfrac{4}{5}=\cfrac{1}{3} → The inverse operation is addition, so add \, \cfrac{4}{5} \, to both sides of the equation.
p-\cfrac{4}{5}+\cfrac{4}{5}=\cfrac{1}{3}+\cfrac{4}{5}
On the left side of the equation: -\cfrac{4}{5}+\cfrac{4}{5}=0, so p-0 is left.
On the right side of the equation: \cfrac{1}{3}+\cfrac{4}{5}=\cfrac{5}{15}+\cfrac{12}{15}=\cfrac{17}{15}
\begin{aligned} & p-\frac{4}{5}+\frac{4}{5}=\frac{1}{3}+\frac{4}{5} \\\\ & p-0=\frac{17}{15} \\\\ & p=\frac{17}{15} \text { or } 1 \frac{2}{15} \end{aligned}
Check:
6. Solve: \cfrac{6 x-9}{3}=1
First, multiply both sides of the equation by 3.
Next, add 9 to both sides.
Finally, divide both sides by 6.
The final answer is x = 2.
You can check the answer by substituting the answer back into the original equation.
\cfrac{6 \, \times \, 2-9}{3}=\cfrac{12-9}{3}=\cfrac{3}{3}=1
Math formulas make work in mathematics more efficient.
While students always spend ample time working to build ideas around a mathematical concept, formalizing it with the use of a formula (like the area of a triangle as A=\cfrac{1}{2} \, b h) allows them to move forward in their understanding and answer more complex questions.
Growing shape patterns start with a shape and grow in a constant way from term to term. Students learn how to write general rules for these patterns. For example, “Start with 2 circles and add 3 circles each time.” This can then be translated into a math equation for the pattern: 2 + 3n = t , where n = the pattern term number and t = the total number of circles.
See also: Shape patterns
Students will learn about the Pythagorean theorem in the 8th grade.
Students start with simple equations and math formulas in elementary school and then progress to solving linear equations and systems of equations in middle school.
In high school, students continue to expand their knowledge by learning more advanced equations such as quadratic equations (including the quadratic formula), exponential equations, and trigonometric equations.
As they move into college level courses such as precalculus and calculus, they will solve rational equations, logarithm equations, parametric equations and differential equations
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