Quick algebra question help?
The difference between the squares of two numbers is 11. Twice the square of the first number increased by the square of the second number is 97. Find the numbers. A planet follows an elliptical path described by 256 x squared plus 16 y squared equals 4096. A comet follows the parabolic path y equals x squared minus 16. Where might the comet intersect the orbiting planet?
Please help with these algebra study questions: Z= ym solve for y?
Lets see what I can do A) -3.2 + -9.2 = -3.2 - 9.2 = -12.4 B) 45/5 =9 C) -4/11 = -0.363636 D and E not my style (I'm English must be different notation) F) 7*-3 = -21 and -5*4 = - 20 gives -21- - 20 =-21 + 20 = -1 G) 8*-9= -72x H) -6 --3 = -6 + 3 = -3 I) Just 45/d (45 over d) J) 6c-c =5c (6 carrots - 1 carrot) K) 8x + 8 +56 ... (8 + 56 = 64) ... so 64/8 =x so (x = 8)
Anyone help on an Algebra question?
to eliminate you must have a positive and a negative of equal value for a variable .3x - .2y = 4 .2x + .5y = 1 you already have a + and a - y, you just need them to have equal values. if i multiply the 1st equation by 5 and the 2nd by 2 the y's will be equal (.3 - .2y = 4)5 and (.2x + .5y = 1)2 you will get 1.5x - 1y = 20 and .4x + 1y = 2 now add the equations together and the y's will eliminate each other you will be left with 1.9x = 22 now divide both sides by 1.9 to isolate x 1.9x/1.9 = 22/1.9 x= 22/1.9 or x = 11.58
What does this algebra question mean, and how do I solve it?
It looks like the main issue is the wording of this question, which I agree is somewhat ambiguous, and also what the symbology,[math]\bigg[ -{1\over 8}, \, \infty \bigg),[/math]as far as being the valid range of the solution, and its use in the equation,[math]y = a\, x^2 + c,[/math]means.The key is the difference between the words “domain” and “range.” The domain is the limit of valid input values that are allowed for the function (the [math]x [/math]values), while “range” (what the question is asking about) is the limit of the valid outputs of the function (the [math]y[/math] values).In this case, the variable [math]x[/math] can presumably vary over its entire range of real values from [math]-\infty[/math] to [math]+\infty[/math]. So we know what [math]x[/math] can do, but we do not know what [math]a[/math] and [math]c[/math] can be. The question is asking you to figure out what the limits are on the [math]a[/math] and [math]c[/math] parameters that will produce the correct range. And what is that range?The lower limit of the range is set as [math]-1/8[/math]. The use of the square bracket on the left (“[“) means that the value of the function ([math]y[/math]) can go down to AND include -1/8. On the high end, what it is saying is that the value can go UP TO positive infinity, but must not include positive infinity (because of the rounded bracket “)”).So here’s how to address this question… Ask yourself what happens as [math]x[/math] becomes much less than or much greater than zero? Obviously [math]x^2[/math] gets big in a positive sense in either direction. That means a positive [math]a[/math] value will cause that term to increase. The “extremum” value of this first term will happen at [math]x=0[/math] and at [math]x=\infty[/math]. The only way to get to infinity in your range will be to make the function infinity at [math]x = \pm\infty[/math]. How do you do that? (Hint: It depends on the value of [math]a[/math].) The answer to this question should tell you how a should be set in your equation.Next, you need to look at what happens when [math]x=0[/math]. In this case the first term goes to zero and your answer becomes [math]c.[/math] Ask yourself what value of [math]c [/math]you could choose to make sure that your range starts at -1/8 when [math]x=0[/math].Once you’ve answered those two questions you will be able to answer the question.
College Algebra help, solve the system: x = y + 4; 3x + 7y = -18?
Here's how you solve for x and y if you didn't have any choices for answers - x=y+4 (Multiply this equation by 3 to get...) 3x=3y + 12 --------- 1 3x+7y = -18 (Do a -7y on both sides of the equation to get...) 3x= -7y-18 ----------2 Now subract equations 1 and 2 3x-3x = 3y + 12 - (-7y-18 ) 0 = 3y+12+7y+18 0 = 10y + 30 (Do a -30 on both sides here to get ... ) -30 = 10y (Divide on both sides to get...) -30/10=y -3 = y Now use value of y in your first equation to find x. x=y+4 x=-3+4 x=1 So, your answer is d. { (1, -3) } Since you have the choices, just substitute each choice in the first equation and check which one fits correctly. That should work pretty easy too. Just wanted for you to know both methods! Good luck!
ALGEBRA HELP! -I need to solve each system by using substitution method?
1st one: 5x-2y=-5; flip 2y to other side 5x=2y-5; divide by 5 x=2/5y-1; put this into 2nd equation for x y-5(2/5y-1)=3 y-2y+5=3 -y=3-5 -y=-2 y=2 put this into 2nd equation for y 2-5x=3 -5x=1 x=-1/5 2nd one 8x-4y=16; divide by 4 2x-y=4; flip 2x to other side -y=4-2x; multiply by -1 y=2x-4; lo and behold this is exactly the same as the 2nd equation - can't be solved with substitution, equations 1 and 2 are the same 3rd one 4x-12y=5; divide by 4 x-3y=5/4; flip 3y to other side x=5/4+3y; substitute this into x in 2nd equation -(5/4+3y)+3y=-1; simplify -5/4-3y+3y=-1; cancel out the -3y and + 3y -5/4=-1; well, how can you have -5/4 ever equal to -1, equations are not the solvable (unless either of the numbers equals infinity)
Intermediate Algebra questions help please?
1. Since x equals 47, we substitute that into the 1st equation and solve for y: 6x+5y = 28 6(47)+5y = 28 282+5y = 28 5y = -254 y = -254/5 Solution ➙ (47, -254/5) 2. To solve by elimination, we add the terms of each equation: x + 6y = 62 -x + 2y = 10 ---------------------- 0 + 8y = 72 The x term was eliminated, so now we can solve for y: 8y = 72 y = 9 Substitute 9 for y in either equation: x+6y = 62 x+6(9) = 62 x+54 = 62 x = 8 Solution ➙ (8,9) 3. As before, we add the terms of the given equations to get: 7r + 3s = 56 Solve that for one of the variables: 7r = 56 - 3s r = 8 - (3/7)s Sub that into one of the equations and solve for the other variable: 5[8-(3/7)s] - 2s = 24 40 - (15/7)s - 2s = 24 40 - (15/7)s - (14/7)s = 24 40 - (29/7)s = 24 -(29/7)s = -16 s = -16(-7/29) s = 112/29 Sub that back into the expression we found for r: r = 8 - (3/7)s r = 8 - (3/7)(112/29) r = 8 - (336/203) r = (232/29) - (48/29) r = 184/29 Solution ➙ (184/29, 112/29) 4. Let t = the amount invested at 10% Let s = the amount invested at 6% The total amount invested was $6300, so we have: s+t = 6300 The total yield was $502, so we have: 0.1t + 0.06s = 502 Solve the first equation for one variable: s = 6300-t Sub that into the second equation and solve for the other variable: 0.1t + 0.06s = 502 0.1t + 0.06(6300-t) = 502 0.1t + 378 - 0.06t = 502 -0.04t = -124 t = 3100 Sub that back into the expression we found for s: s = 6300 - t s = 6300 - 3100 s = 3200 ➙ So $3,100 was invested at 10% and $3,200 was invested at 6%.
How do I solve this algebra problem?
Finding the X and Y InterceptsFinding the y-intercept is easy, just plug in [math] x = 0 [/math].[math] 2(0)^2 + 8(0) - 10 = -10[/math]Thus the y-intercept is -10.Finding the x-intercept is equivalent to solving [math] 2x^2 + 8x - 10 = 0 [/math] for x. This can be done as follows[math]\begin{align}2x^2 + 8x - 10 & = 0 \\2(x^2 + 4x - 5) & = 0 \\x^2 + 4x - 5 = 0 \\(x + 5)(x - 1) = 0\end{align}[/math]This equation has two solutions, [math] x = -5,\ 1 [/math]. Thus the two x-intercepts of [math] 2x^2 + 8x - 10 [/math] are at -5 and 1.Finding the VertexTo find the vertex, we have to put the equation in the form [math] f(x) = a(x - k)^2 + h [/math], in which case the vertex will be the point [math] (k, h) [/math].[math]\begin{align}2x^2 + 8x - 10 & =2(x^2 + 4x - 5) \\& = 2(x^2 + 4x + 4 - 4 - 5) \\& = 2\left((x + 2)^2 - 9\right) \\& = 2(x + 2)^2 - 18\end{align}[/math]From this we see that [math] k = -2 [/math], and [math] h = -18 [/math]. Thus the vertex is the point [math] (-2, -18) [/math].Finding the RangeTo find the range, lets think about what our function looks like. We'll think about it in vertex form ([math] f(x) = 2(x + 2)^2 - 18 [/math]). We know that [math] (x + 2)^2 [/math] is always positive for all values of [math] x [/math], thus the minimum (x + 2)^2 will ever be is when [math] x = -2 [/math], in which case [math] (x + 2)^2 = 0 [/math].So if the minimum value [math] (x + 2)^2 [/math] can ever be is 0, then what is the minimum value [math] 2(x + 2)^2 - 18 [/math] can ever be? Well, when we plug in [math] x = -2 [/math], we get -18, so the minimum of our graph is [math] y = -18 [/math].How do we know it's a minimum? [math] (x + 2)^2 [/math] is always positive, so that means it will always be greater than its minimum value. Thus, since the minimum is [math] -18 [/math], and [math] 2(x + 2)^2 - 18 [/math] grows without bound (i.e. it gets infinitely large as x gets infinitely large), we can conclude that the range of the function is [math] [-18, \infty) [/math].
How do I solve substitution problems in algebra?
In substitution method, find value of one variable in terms of other variable from one equation. Then substitute the value of that variable in other equation. Then you get equation in one variable, which can be solved. After getting value of one variable, you can substitute in any equation to get the value of other variable. Bit difficult to understand. OK, I shall explain by simple example:Suppose you want to solve the equations: x +2y = 4 equation I and 3x + y = 7 equation IIFrom equation I : x = 4 – 2y now substitute value of x in equation II:3*(4-2y) + y = 7 therefore12 – 6y + y = 7 therefore 5 = 5y therefore y =1.Substituting value of y =1 in equation x = 4 – 2y , therefore x = 4 – 2 * 2 , therefore x =1. Hope I made it clear. The basic principle can be applied in other examples also. Best regards. --- Sanjay Chakradeo