Solve the system of equations algebraically. Show all of your steps.
y=x^2+2x
**y=x^2+2x
(2/2)^2=1
1=answer**
y=3x+20
**y=3x+20
(3/2)^2=2.25
2.25=answer**
am i right or wrong, if im wrong could someone show me the steps
the system of equations is
y=x^2+2x
y=3x+20
You want to find a value of x that makes both equations true. So, since the expressions must be equal,
x^2 + 2x = 3x + 20
x^2 - x - 20 = 0
(x-5)(x+4) = 0
x = 5 or -4
check:
5^2 + 2*5 = 35
3*5 + 20 = 35
(-4)^2 + 2(-4) = 16-8 = 8
3(-4)+20 = -12+20 = 8
So, yes, both values are solutions to both equations
First off that's the test answers for Unit 5 Lesson 10, to add the answers listed are also incorrect
Well, Clown Bot is not very good at solving equations algebraically, but I can certainly give it a try! Let's solve the system of equations step by step.
Equation 1: y = x^2 + 2x
Equation 2: y = 3x + 20
To solve this system of equations, we can set the right sides of both equations equal to each other:
x^2 + 2x = 3x + 20
Now let's rearrange the equation to bring all terms to one side:
x^2 + 2x - 3x - 20 = 0
Simplifying further:
x^2 - x - 20 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. Since factoring might be a bit tricky in this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = -1, and c = -20.
x = (-(-1) ± √((-1)^2 - 4(1)(-20))) / (2(1))
x = (1 ± √(1 + 80)) / 2
x = (1 ± √81) / 2
x = (1 ± 9) / 2
Now we have two possible solutions for x:
x = (1 + 9) / 2 = 10 / 2 = 5
x = (1 - 9) / 2 = -8 / 2 = -4
To find the corresponding y-values, we can substitute these x-values into either of the original equations. Let's use the first equation:
For x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35
For x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
So, the solution to the system of equations is (x, y) = (5, 35) and (-4, 8).
And remember, Clown Bot is here to lighten the mood, even if the math gets a little heavy!
To solve the given system of equations algebraically, we need to find the values of x and y that satisfy both equations simultaneously.
Let's solve the system step by step.
Equation 1: y = x^2 + 2x
Equation 2: y = 3x + 20
To begin, we can set the two equations equal to each other since they both equal y:
x^2 + 2x = 3x + 20
Now, let's move all terms to one side to set the equation equal to zero:
x^2 + 2x - 3x - 20 = 0
Combining like terms, the equation simplifies to:
x^2 - x - 20 = 0
Next, we can factor the quadratic equation, if possible. Let's find two numbers that multiply to -20 and add up to -1:
The factors are -5 and 4, as -5 * 4 = -20 and -5 + 4 = -1.
Therefore, we can rewrite the equation as:
(x - 5)(x + 4) = 0
Now, we can set each factor equal to zero and solve for x:
Case 1: x - 5 = 0
Adding 5 to both sides, we get:
x = 5
Case 2: x + 4 = 0
Subtracting 4 from both sides, we get:
x = -4
Hence, we have found two possible values for x: x = 5 and x = -4.
To find the corresponding y-values, we substitute these x-values back into either of the original equations.
For x = 5:
Using Equation 1: y = x^2 + 2x
y = 5^2 + 2(5)
y = 25 + 10
y = 35
For x = -4:
Using Equation 1: y = x^2 + 2x
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
So, the solution to the system of equations is x = 5, y = 35 and x = -4, y = 8.
1.b
2.b
3.d
4.c
5.b
6.a
7.a
8.b
9.b
10.b
11.c
12.a
13.b
14.c
15.c
16.c
17.b
18.b
19.d
20.b
21.c