Word problems and Exam view on algebra 2
Word problems and Exam view on algebra 2
Word problems are story problems that narrate a situation and ask us to find the value of the unknowns. Algebra 2 includes topics like linear equations, quadratic equations and solving them. First we have to understand the given word problem , interpret the data from the word problem and then form an algebraic expression or equation according to given data and then solve the equation to get the solution for given problem. In this article let us learn how to frame the linear and quadratic equations from the given word problems on algebra 2 with the steps and as well as how to solve them.
Learning word problems on algebra 2:
There are 2 example problems explained in this article. One word problem is on linear equation and the other is on quadratic equation. Linear equation can have one or more than one variables. Any equation of the form ax+by+c = 0 is called as linear equation with two variables x and y. Any equation of the form ax2 + bx + c = 0 is called as quadratic equation.
Exam view Algebra 2 is a division of mathematics that substitutes letters for numbers. In this division of mathematics, we use letters like a, b, x and y to denote numbers.
The stage of addition, subtraction, multiplication, division or extraction of roots on these symbols and real numbers performed are called algebraic expression in exam view algebra 2.
An algebraic equation represents a scale, what is finished on one side of the scale with a number is also completed on the other side of the scale.
In this article, we shall learn about the exam view algebra 2. Also we shall solve sample problems based on exam view algebra 2.
Exam view algebra 2 problems:
Example 1:
Find the sum of 7x4 – 6x2 + 7x + 28 and 1x + 2x3 – 3x2 – 24.
Solution:
By means properties of real numbers, we can express
(7x4 – 6x2 + 7x + 28) + (2x3 – 3x2 + 1x – 24) = 7x4 + 2x3 – 6x2 – 3x2 + 7x + 1x + 28 – 24
= 7x4 + 2x3 – 9x2 + 7x + 1x + 28 – 24
= 7x4 + 2x3 – 9x2 + 8x + 4
So the final result is 7x4 + 2x3 – 9x2 + 8x + 4
Example 2:
Subtract 35x3 + 5x2 – 6 from 32x3 – x2 – 1.
Solution:
Using properties, we have
(32x3 + x2 –1) – (35x3 + 5x2 – 6) = 32x3 + x2 – 1 – 35x3 -5x2 + 6
= 32x3 – 35x3 +x2 -5x2 +6 - 1
= (32x3 – 35x3) + (-5x2 + x2) + (6-1)
= – 3x3-4x2 + 5.
So the final result is – 3x3-4x2 + 5.
Example 3:
Factorize x2 – xy – x + y.
Solution:
The expressions do not have a general factor. However, we detect that the terms can be combined as follows:
x2 – xy – x + y = (x2 – xy) – (x – y)
= x(x – y) –1(x – y)
= (x – y) (x –1)
So the final result is (x – y) (x – 1).
Exam view algebra 2 problems with line equations:
Example 4:
Solve the linear equation:
x + 2y = 6
x - 4y =12
Solution:
Using substitution method, assign the above equations are 1 and 2
equation 1 => x + 2y = 6
x = 6 - 2y
Substitute the value of x in 2,
6 - 2y - 4y = 12
6 - 6y = 12
-6y = 12 - 6
-6y = 6
y = -1
Substitute the value of y in equation 1 or 2.
=> x + 2(-1) = 6
x - 2 = 6
x = 6 + 2
x = 8
So the final result is x = 8, y = -1
Word problems on algebra 2:
Ex 1: Two buses are lets started from bus stand same time.Two buses are going on opposite direction. One bus goes 20 miles per second faster than the other. After 6 hours both are 150 miles apart. How fast both are going?
Sol:
Step 1: Find the given data from the word problem:
Time = 6 hours
Distance apart from the both bus = 150miles
Step 2:
Let x be the speed of the bus 1. So, speed of the bus 2 is x+20
Step 3: Find which formula is useful.
Formula for finding the Distance =Speed * Time
Speed=Distance/Time
Distance covered by bus 1 = 6x
Distance covered by bus 2 =(x+20)*6
Step 4: Distance between the two buses = 150 miles
6(x+20)+6x=150
6x+120+6x=150
12x+120=150( subtract both side by 120)
12x+120-120= 150-120
12x= 150-120
12x=30( Divide both side by 12)
X=30 / 12
X=2.5
Step 5: Speed of bus 1 = 2.5 miles per second
Speed of bus 2 = 22.5 miles per second.
Ex 2: Find the measurement of a rectangular room whose length is 10 more than the width and area of the rectangular room is 48 square units.
Sol:
Step 1: Find the given data from the word problem:
Length of the rectangular room = 10 more than the width
Area of the rectangle=48 sq units
Step 2: Let l, w be the length and width of the rectangular room.
So, l = w+10
Step 3: Area of the rectangle=Length* Width
Length * width = (w+10)* w = 48
w(w+10)=48
w2+10w =48
w2+10w-48=0
Step 4: Use quadratic formula and find the value of w
w = `(-b+-sqrt(b^(2)-4ac))/(2a)`
From equation a=1,b=10,c=-48
w = `(-10+-sqrt(10^(2)-4(1)(-48)))/(2(1))`
w = 3.5,-13.4
Step 5: Width of the rectangular room =3.5 units
Length of the rectangular room = x+10 = 3.5+10 = 13.5 units
Practice the following word problems on algebra 2:
Two trains are started from the railway station at the same time. Two trains are going on opposite direction. One train goes 25 miles per second faster than the other. After 3 hours both are 300 miles apart. Calculate the speed of the two trains.
Find the measurement of a rectangular room whose length is 4 more than the width and perimeter of the rectangular room is 48 square units.
Word problems are story problems that narrate a situation and ask us to find the value of the unknowns. Algebra 2 includes topics like linear equations, quadratic equations and solving them. First we have to understand the given word problem , interpret the data from the word problem and then form an algebraic expression or equation according to given data and then solve the equation to get the solution for given problem. In this article let us learn how to frame the linear and quadratic equations from the given word problems on algebra 2 with the steps and as well as how to solve them.
Learning word problems on algebra 2:
There are 2 example problems explained in this article. One word problem is on linear equation and the other is on quadratic equation. Linear equation can have one or more than one variables. Any equation of the form ax+by+c = 0 is called as linear equation with two variables x and y. Any equation of the form ax2 + bx + c = 0 is called as quadratic equation.
Exam view Algebra 2 is a division of mathematics that substitutes letters for numbers. In this division of mathematics, we use letters like a, b, x and y to denote numbers.
The stage of addition, subtraction, multiplication, division or extraction of roots on these symbols and real numbers performed are called algebraic expression in exam view algebra 2.
An algebraic equation represents a scale, what is finished on one side of the scale with a number is also completed on the other side of the scale.
In this article, we shall learn about the exam view algebra 2. Also we shall solve sample problems based on exam view algebra 2.
Exam view algebra 2 problems:
Example 1:
Find the sum of 7x4 – 6x2 + 7x + 28 and 1x + 2x3 – 3x2 – 24.
Solution:
By means properties of real numbers, we can express
(7x4 – 6x2 + 7x + 28) + (2x3 – 3x2 + 1x – 24) = 7x4 + 2x3 – 6x2 – 3x2 + 7x + 1x + 28 – 24
= 7x4 + 2x3 – 9x2 + 7x + 1x + 28 – 24
= 7x4 + 2x3 – 9x2 + 8x + 4
So the final result is 7x4 + 2x3 – 9x2 + 8x + 4
Example 2:
Subtract 35x3 + 5x2 – 6 from 32x3 – x2 – 1.
Solution:
Using properties, we have
(32x3 + x2 –1) – (35x3 + 5x2 – 6) = 32x3 + x2 – 1 – 35x3 -5x2 + 6
= 32x3 – 35x3 +x2 -5x2 +6 - 1
= (32x3 – 35x3) + (-5x2 + x2) + (6-1)
= – 3x3-4x2 + 5.
So the final result is – 3x3-4x2 + 5.
Example 3:
Factorize x2 – xy – x + y.
Solution:
The expressions do not have a general factor. However, we detect that the terms can be combined as follows:
x2 – xy – x + y = (x2 – xy) – (x – y)
= x(x – y) –1(x – y)
= (x – y) (x –1)
So the final result is (x – y) (x – 1).
Exam view algebra 2 problems with line equations:
Example 4:
Solve the linear equation:
x + 2y = 6
x - 4y =12
Solution:
Using substitution method, assign the above equations are 1 and 2
equation 1 => x + 2y = 6
x = 6 - 2y
Substitute the value of x in 2,
6 - 2y - 4y = 12
6 - 6y = 12
-6y = 12 - 6
-6y = 6
y = -1
Substitute the value of y in equation 1 or 2.
=> x + 2(-1) = 6
x - 2 = 6
x = 6 + 2
x = 8
So the final result is x = 8, y = -1
Word problems on algebra 2:
Ex 1: Two buses are lets started from bus stand same time.Two buses are going on opposite direction. One bus goes 20 miles per second faster than the other. After 6 hours both are 150 miles apart. How fast both are going?
Sol:
Step 1: Find the given data from the word problem:
Time = 6 hours
Distance apart from the both bus = 150miles
Step 2:
Let x be the speed of the bus 1. So, speed of the bus 2 is x+20
Step 3: Find which formula is useful.
Formula for finding the Distance =Speed * Time
Speed=Distance/Time
Distance covered by bus 1 = 6x
Distance covered by bus 2 =(x+20)*6
Step 4: Distance between the two buses = 150 miles
6(x+20)+6x=150
6x+120+6x=150
12x+120=150( subtract both side by 120)
12x+120-120= 150-120
12x= 150-120
12x=30( Divide both side by 12)
X=30 / 12
X=2.5
Step 5: Speed of bus 1 = 2.5 miles per second
Speed of bus 2 = 22.5 miles per second.
Ex 2: Find the measurement of a rectangular room whose length is 10 more than the width and area of the rectangular room is 48 square units.
Sol:
Step 1: Find the given data from the word problem:
Length of the rectangular room = 10 more than the width
Area of the rectangle=48 sq units
Step 2: Let l, w be the length and width of the rectangular room.
So, l = w+10
Step 3: Area of the rectangle=Length* Width
Length * width = (w+10)* w = 48
w(w+10)=48
w2+10w =48
w2+10w-48=0
Step 4: Use quadratic formula and find the value of w
w = `(-b+-sqrt(b^(2)-4ac))/(2a)`
From equation a=1,b=10,c=-48
w = `(-10+-sqrt(10^(2)-4(1)(-48)))/(2(1))`
w = 3.5,-13.4
Step 5: Width of the rectangular room =3.5 units
Length of the rectangular room = x+10 = 3.5+10 = 13.5 units
Practice the following word problems on algebra 2:
Two trains are started from the railway station at the same time. Two trains are going on opposite direction. One train goes 25 miles per second faster than the other. After 3 hours both are 300 miles apart. Calculate the speed of the two trains.
Find the measurement of a rectangular room whose length is 4 more than the width and perimeter of the rectangular room is 48 square units.
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