Solving Linear Equations: Expanding Brackets, Subbing Values

In summary: Yeah, my calculator is being a retard so i made the scale the same as my mate's one which is the same model, only newer. Thanks for all your help tiny tim:cool: Now I can finish off my assignment:rofl:
  • #1
TheAkuma
53
0
Okay, so I expanded the brackets and simplified the equation from (x+4)(x-4)(x-4) to equal
x3-4x2-16x+64 and I needed to sub those values into this equation y=ax3+bx2+cx-d where a=x3 b=-4x2 c=-16 d=64 which were the values I got from the first equation. I need confirmation If it would be y=x6[/SUP]-4x4-16x2-64
I dunno, it just doesn't seem right.
 
Physics news on Phys.org
  • #2
TheAkuma said:
… x3-4x2-16x+64 and I needed to sub those values into this equation y=ax3+bx2+cx-d where a=x3 b=-4x2 c=-16 d=64 which were the values I got from the first equation.

Hi TheAkuma! :smile:

Are you sure you don't mean a = 1, b = -4, c = -16, d = 64? :confused:

What is the whole problem?
 
  • #3
tiny-tim said:
Hi TheAkuma! :smile:

Are you sure you don't mean a = 1, b = -4, c = -16, d = 64? :confused:

What is the whole problem?

I honestly don't know:confused: did you just ignore the 'x'?
 
  • #4
TheAkuma said:
I honestly don't know:confused: did you just ignore the 'x'?

Well, if x3-4x2-16x+64 is the same as y=ax3+bx2+cx-d, then a b c and d can't have any x in them.

Or am I misunderstanding the question?
 
  • #5
tiny-tim said:
Well, if x3-4x2-16x+64 is the same as y=ax3+bx2+cx-d, then a b c and d can't have any x in them.

Or am I misunderstanding the question?

What am I smoking? sorry about that. The question is already solved. I just had to expand the brackets into the form of y=ax3+bx2+cx-d which when solved is y=x3-4x2-16x-64 where 'a' in that equation would be 1 and 'b' would be '-4' and so on. Sorry about that
 
  • #6
TheAkuma said:
What am I smoking? sorry about that.

he he :biggrin:

that's ok! :smile:​
 
  • #7
uh oh. When I try to graph it on my calculator it comes up blank. Even though I spent one whole day trying to fix my graphics calculator. And when I try to do it manually, the parabola or whatever the hell its called is waay at the bottom of my graph. What am I doing wrong?
 
  • #8
TheAkuma said:
uh oh. When I try to graph it on my calculator it comes up blank. Even though I spent one whole day trying to fix my graphics calculator. And when I try to do it manually, the parabola or whatever the hell its called is waay at the bottom of my graph. What am I doing wrong?

y = (x+4)(x-4)(x-4) is a cubic …

it goes up, down, and up …

it cuts the x-axis at x = -4, goes up to about 100, comes down and touches the x-axis at x = 4, and goes up again …

is that why it's off the graph?

Try dividing everything by 10. :smile:
 
  • #9
tiny-tim said:
y = (x+4)(x-4)(x-4) is a cubic …

it goes up, down, and up …

it cuts the x-axis at x = -4, goes up to about 100, comes down and touches the x-axis at x = 4, and goes up again …

is that why it's off the graph?

Try dividing everything by 10. :smile:

Yeah, my calculator is being a retard so i made the scale the same as my mate's one which is the same model, only newer. Thanks for all your help tiny tim:cool: Now I can finish off my assignment:rofl:
 

1. What are linear equations?

Linear equations are mathematical expressions that contain only constants and variables raised to the first power. They can be represented by a straight line on a graph and follow the form y = mx + b, where m is the slope and b is the y-intercept.

2. How do you expand brackets in a linear equation?

To expand brackets in a linear equation, you need to distribute the value outside the brackets to each term inside the brackets. This means multiplying the value by each term inside the brackets. For example, in the equation 3(x + 2), you would distribute the 3 to both x and 2, resulting in 3x + 6.

3. What is the process for substituting values in a linear equation?

Substituting values in a linear equation involves replacing the variable with a specific numerical value and then solving the resulting equation. This can be done by plugging in the value for the variable and then following the order of operations to simplify the equation.

4. Why is it important to solve linear equations?

Solving linear equations is important because it allows us to find the value of a variable in an equation and helps us solve real-world problems. It also helps us understand the relationship between variables and how changing one variable can affect the other.

5. What are some common mistakes when solving linear equations?

Some common mistakes when solving linear equations include forgetting to apply the distributive property, making errors in simplifying expressions, and incorrectly combining like terms. It is important to carefully follow the order of operations and double-check your work to avoid these mistakes.

Similar threads

I Canonical Form for quadratic equations *with* linear terms
    • Linear and Abstract Algebra
Replies
4
Views
2K
I Solving a complex linear system with parameters
    • Linear and Abstract Algebra
Replies
11
Views
1K
A Selected solution of linear equations
    • Linear and Abstract Algebra
Replies
8
Views
1K
Question about this technique for solving simultaneous equations
    • Precalculus Mathematics Homework Help
Replies
4
Views
612
MHB Solving Linear Equations: $Ax=b$ and Rank of A
    • Linear and Abstract Algebra
Replies
9
Views
2K
MHB Solving a system of linear equations with one unknown value
    • Linear and Abstract Algebra
Replies
2
Views
2K
B What Does 'Simultaneous' Mean in Linear Equations?
    • Linear and Abstract Algebra
Replies
1
Views
1K
I DSP: Recurrence Relations in a Linear Algebra Equation
    • Linear and Abstract Algebra
Replies
1
Views
2K
I How to find the value of a constant experimentally?
    • Other Physics Topics
Replies
8
Views
2K
I Fundamental matrix of a second order 2x2 system of ODEs
    • Differential Equations
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top