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# how do you solve 5-(t+3)=-1-2 (t-3)

I need to know how to solve this, step by step. Help, anyone? Note: I said step by step, so I dont want the answer. I want to know how to get the answer.

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### How do you solve 3t(t+5)-t^2=2t^2+4t-1? | Socratic

... (t+5)-t^2 = 3t^2+15t-t^2 = 2t^2+15t So the equation becomes: ... SOCRATIC Subjects . Science Anatomy ... How do you solve #3t(t+5)-t^2=2t^2+4t-1#?

### How would you evaluate : [math] 2^{5/2}-2^{3/2 ... - Quora

How would you evaluate : [math] 2^{5/2}-2^{3/2}= [/math]? ... 2^{5/2} - 2^{3/2} = 2^{3/2} (2 - 1) = 2^{3/2} ... How do you solve [math]x^2=2^x[/math] ...

### Solving Exponential Equations from the Definition | Purplemath

Demonstrates how to solve exponential equations by using the definition of exponentials, ... Solve 5 x = 5 3; Since the bases ("5" in each case) are the same, ...

### how do i solve this math problem 4 2/5 * 1 1/4 - 1 3/5 / 1 1 ...

Resources / Answers / how do i solve this math ... Ask a question. 0 0 how do i solve ... how do i solve this math problem 4 2/5 * 1 1/4 ...

### How do you solve for t in 2/7(t+2/3)=1/5(t-2/3)? | Socratic

... = 1/5 (t-2/3) Multiplying , we get (2/7) ... SOCRATIC Subjects . Science Anatomy ... How do you solve for t in #2/7(t+2/3)=1/5 ...

### 6^2 ÷ 2(3) + 4 = ? The Correct Answer - YouTube

Apr 28, 2015 · 6^2 ÷ 2(3) + 4 = ? The Correct Answer ... You may also like my video on 6 - 1 x 0 + 2 ÷ 2 which explains the order of ... Can You Solve The Viral 1 + 4 ...

### 5 Ways to Solve for X - wikiHow - How to do anything

Just divide 3x and 9 by 3, the x term coefficient, to solve for x. 3x/3 = x and 3/3 = 1, so you're left with x = 1. 5. Check your work. To check your work, ...

### 3 Ways to Solve Exponents - wikiHow

How to Solve Exponents. ... 10. But how do I solve when it's 5.5? Atheia. The Richter scale is a log 10 scale. This means that when you calculate ratios, ...

## Suggested Questions And Answer :

### How do I solve y=-x-4

How do I solve y=-x-4 I need to be able to plot these points on a graph but i don't know how to solve the problem.   make an x & y axis. ( like a cross one line goes vertical an the other horizontal) put equal dimention for x & y line From center line moving to the left is negative line moving down is negative Just assign values for x & y then locate it at the x & y axis.                                                                                                                                                            SOLVE Y                                                                                        POINTS X Y 1 0 SOLVE Y 2 SOLVE X 0 3 1 SOLVE Y 4 SOLVE X 1 5 2 SOLVE Y 6 SOLVE X 2 7 -1 8 SOLVE X -1 9 -2 SOLVE Y 10 SOLVE X -2 11 3 SOLVE Y y=-x-4 sample: in point 1 => if  x = 0 y = - (0) - 4 y = - 4 in point 2 => if y = 0 0 = - x - 4 x = - 4 in point 7 => if x=-1 y = -x -4 y =-(-1) -4 y = +1 - 4 y = - 3 at point 8 => if y = -1 -1 = -x -4 x = -4 + 1 x = - 3 - 3 POINTS X Y 1 0 - 4 2 - 4 0 3 1 - 5 4 - 5 1 5 2 - 6 6 - 6 2 7 -1 8 - 3 -1 9 -2 - 2 10 - 2 -2 11 3 - 7 if you have done the 1 to 4 then you can plot the points from 1 to 11 and there is your graph.

### how to solve for x with fractions

Simplifying x3 + 3x2 + -4x = 0 Reorder the terms: -4x + 3x2 + x3 = 0 Solving -4x + 3x2 + x3 = 0 Solving for variable 'x'. Factor out the Greatest Common Factor (GCF), 'x'. x(-4 + 3x + x2) = 0 Factor a trinomial. x((-4 + -1x)(1 + -1x)) = 0 Subproblem 1 Set the factor 'x' equal to zero and attempt to solve: Simplifying x = 0 Solving x = 0 Move all terms containing x to the left, all other terms to the right. Simplifying x = 0 Subproblem 2 Set the factor '(-4 + -1x)' equal to zero and attempt to solve: Simplifying -4 + -1x = 0 Solving -4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '4' to each side of the equation. -4 + 4 + -1x = 0 + 4 Combine like terms: -4 + 4 = 0 0 + -1x = 0 + 4 -1x = 0 + 4 Combine like terms: 0 + 4 = 4 -1x = 4 Divide each side by '-1'. x = -4 Simplifying x = -4 Subproblem 3 Set the factor '(1 + -1x)' equal to zero and attempt to solve: Simplifying 1 + -1x = 0 Solving 1 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + -1x = 0 + -1 Combine like terms: 1 + -1 = 0 0 + -1x = 0 + -1 -1x = 0 + -1 Combine like terms: 0 + -1 = -1 -1x = -1 Divide each side by '-1'. x = 1 Simplifying x = 1 Solution x = {0, -4, 1}

### How do you solve this problem? solve for d: n= m+3d/4

Problem: How do you solve this problem? solve for d: n= m+3d/4 I need help solving this and seeing how it was solved since I am having trouble figuring out what to do and how to solve it thanks. n = m + 3d/4 n - m = (m + 3d/4) - m n - m = 3d/4 4(n - m) = (3d/4)4 4(n - m) = 3d 4(n - m)/3 = 3d/3 4/3 (n - m) = d Answer: d = 4/3 (n - m)

### Solve, using linear combination. 3x + y = 4 2x + y = 5

line 1: 3x+y=4    or y=4-3x line 2: 2x+y=5 line 1 -line2...(3x-2x)+(y-y)=4-5 or x=-1 y=4-3x=4+3=7

### how can i solve these equations?

how can i solve these equations? x + 3y – z = 2 x – 2y + 3z = 7 x + 2y – 5z = –21 We eliminate one of the unknowns (x, y or z, take your choice), leaving two unknowns. Then, we eliminate a second one, giving us an equation with only one of the unknowns. Solve for that and plug that value into one of the equations to solve for a second unknown. Finally plug both of those values into an equation to solve for the third unknown. It sounds complicated, but if you follow a logical sequence, the problem solves itself. 1) x + 3y – z = 2 2) x – 2y + 3z = 7 3) x + 2y – 5z = –21 If we subtract equation 3 from equation 2, we eliminate the x.    x – 2y + 3z =     7 -(x + 2y – 5z = –21) ------------------------     - 4y  + 8z =   28 4) -4y + 8z = 28 Subtract equation 1 from equation 2, eliminating the x again.    x – 2y + 3z = 7 -(x + 3y –   z = 2) ----------------------      - 5y +  4z = 5 5) -5y + 4z = 5 You now have two equations with only a y and a z. The easiest step now is to eliminate the z. Multiply equation 5 by 2. 2 * (-5y + 4z) = 5 * 2 6) -10y + 8z = 10 Subtract equation 6 from equation 4, eliminating the z.     -4y + 8z = 28 -(-10y + 8z = 10) ---------------------      6y      =   18 6y = 18 y = 3  <<<<<<<<<<<<<<<<<<<<< Plug that into equation 5 to solve for z. -5y + 4z = 5 -5(3) + 4z = 5 -15 + 4z = 5 4z = 20 z = 5  <<<<<<<<<<<<<<<<<<<<< Plug the values of y and z into equation 1 to solve for x. x + 3y – z = 2 x + 3(3) – 5 = 2 x + 9 - 5 = 2 x + 4 = 2 x = -2  <<<<<<<<<<<<<<<<<<<<< Always check the answers by plugging all three values into one of the original equations. Using all three would be even better. Equation 2: x – 2y + 3z = 7 (-2) – 2(3) + 3(5) = 7 -2 - 6 + 15 = 7 -8 + 15 = 7 7 = 7 Equation 3: x + 2y – 5z = –21 (-2) + 2(3) – 5(5) = –21 -2 + 6 - 25 = -21 6 - 27 = -21 -21 = -21 Answer: x = -2, y = 3, z = 5

### Inequalities

9x+6=51 9x=51-6 9x=45 x=5

### how do you solve 3/x+5

how do you solve 3/x+5 I'm confused on how to solve this problem All you have is an expression. There is nothing to solve until you make it an equation. 3/x + 5 = -8 3/x + 5 = 14 3/x + 5 = 0 3/x + 5 = 38.67 3/x + 5 = -91 3/x + 5 = 43 Those can be solved, because they are complete. Each one will have a different value for x because of the value on the right side of the equation. You have not given us anything to solve.

### x - z = -3, y + z = 9, -2x + 3y +5z = 33

Problem: x - z = -3, y + z = 9, -2x + 3y +5z = 33 1) x - z = -3 2) y + z = 9 3) -2x + 3y + 5z = 33 Add equation 2 to equation 1.      x     - z = -3 +(    y + z =  9) ------------------   x + y     = 6 4) x + y = 6 Multiply equation 2 by 5. 5(y + z) = 9 * 5 5) 5y + 5z = 45 Subtract equation 3 from equation 5.             5y + 5z = 45 -(-2x + 3y + 5z = 33) --------------------------     2x + 2y        = 12 6) 2x + 2y = 12 Multiply equation 4 by 2. 2(x + y) = 6 * 2 7) 2x + 2y = 12 Subtract equation 7 from equation 6.    2x + 2y = 12 -(2x + 2y = 12) --------------------   0x      = 0      We are solving for x, not y x = 0    <<<<<<<<<<<<<<<<<<< Use equation 4 to solve for y. x + y = 6 0 + y = 6 y = 6    <<<<<<<<<<<<<<<<<<< Use equation 3 to solve for z. -2x + 3y + 5z = 33 -2(0) + 3(6) + 5z = 33 0 + 18 + 5z = 33 5z = 15 z = 3    <<<<<<<<<<<<<<<<<<< Check the values. 1) x - z = -3    0 - 3 = -3    -3 = -3 2) y + z = 9    6 + 3 = 9    9 = 9 3) -2x + 3y + 5z = 33    -2(0) + 3(6) + 5(3) = 33    0 + 18 + 15 = 33    33 = 33 Those numbers work..... ------------- What if we had solved for y instead of x here:    2x + 2y = 12 -(2x + 2y = 12) -------------------        0y = 0      We are solving for y, not x y = 0    <<<<<<<<<<<<<<<<<<< Use equation 4 to solve for x. x + y = 6 x + 0 = 6 x = 6    <<<<<<<<<<<<<<<<<<< Use equation 3 to solve for z. -2x + 3y + 5z = 33 -2(6) + 3(0) + 5z = 33 -12 + 0 + 5z = 33 5z = 45 z = 9    <<<<<<<<<<<<<<<<<<< Check the values. 1) x - z = -3    6 - 9 = -3    -3 = -3 2) y + z = 9    0 + 9 = 9    9 = 9 3) -2x + 3y + 5z = 33    -2(6) + 3(0) + 5(9) = 33    -12 + 0 + 45 = 33    33 = 33 Those numbers work, too. That complicates the solution. Answer 1: x = 0, y = 6, z = 3 Answer 2: x = 6, y = 0, z = 9

### how to solve by substitution

how to solve by substitution .3x+3y=1.2(100) "Substitution" implies there are TWO equations in two unknowns. You solve one of them for one of the unknowns in terms of the other, e.g., solve for y in terms of x. THEN, you substitute that value into the other equation where the y is, and you have one equation with only one unknown, in this case x. You solve for x and substutute that into the first equation to solve for y. Having only one equation makes it impossible to solve for either unknown.

### 2x+y=9 x-2z=-3 2y+3z=15

2x+y=9 x-2z=-3 2y+3z=15 Forgot how to do This. Problems Solving systems of this type requires eliminating all but one unknown so you can solve for that unknown. Then, plug that value into one of the equations to solve for another of the unknowns. Finally, plug both of those values into an equation to solve for the last unknown. 1) 2x + y = 9 2) x - 2z = -3 3) 2y + 3z = 15 Multiply equation 2 by 2 so we can subtract it from equation 1. 2 * (x - 2z) = -3 * 2 4) 2x - 4z = -6   2x + y       =  9 -(2x     - 4z = -6) ---------------------         y + 4z = 15 5) y + 4z = 15 We now have two equations with y and z. Multiply equation 5 by 2... 2 * (y + 4z) = 15 * 2 6) 2y + 8z = 30 ...and subtract equation 6 from equation 3. Then, solve for z.   2y +  3z = 15 -(2y + 8z = 30) -------------------        - 5z = -15 -5z = -15 z = 3     <<<<<<<<<<<<<<<<<<<< Plug that into equation 3 to solve for y. 2y + 3z = 15 2y + 3(3) = 15 2y + 9 = 15 2y = 6 y = 3     <<<<<<<<<<<<<<<<<<<< Also, plug the z value into equation 2 to solve for x. x - 2z = -3 x - 2(3) = -3 x - 6 = -3 x = 3     <<<<<<<<<<<<<<<<<<<< Plug the values for x and y into equation 1 to verify that they are correct. 2x + y = 9 2(3) + 3 = 9 6 + 3 = 9 9 = 9 Usually, all three equations have all three unknowns, requiring more manipulation and more elimination, but the process is the same. For this problem, x = 3, y = 3 and z = 3.