Guide :

# 2x+4y+3z=-11 4x-3y+4z=34 x+2y-4z=-33 solve the system

Solve the system using the elimination method

## Research, Knowledge and Information :

### 4x+2y-3z=7 2x+4y+z=35 6x+2y-2z=20 solve the system

Show all your work 3x-2y+2z= 30 -x+3y-4z= -33 2x-4y+3z= 42 ... Solve each system. x-4y-6z=-34 -7x+5y-7z=-24 z=5x-7y+18 ... Then classify the system? y= -x -11 4x -3y ...

### Use Matrices To Solve The System. 1) X - 2y - 3z =... | Chegg.com

Answer to Use matrices to solve the system. 1) x - 2y - 3z = -1 2x + y + z = 6 x + 3y - 2z = 13 3) 5x + 2y -z = -7 x ... 4x - 2y + z = 5. 3x + y ...

### 3x-2y+2z=30 -x+3y-4z=-33 2x-4y+3z=42 - Wyzant Answers

3x-2y+2z=30 -x+3y-4z=-33 2x-4y+3z=42. solve for the following system show your work. ... (eq.#2) -x + 3y - 4z = -33 (eq.#3) 2x ...

### solve the following system 2x+3x-z=1 3x+y+2z=12 ... - OpenStudy

solve the following system 2x+3x-z=1 3x+y+2z=12 x+2y-3z=-5 a) ... 2x+3y-z=1 3x+y+2z=12 x+2y-3z=-5. ... - -> x=3 11(3) + 7y = 26 33 + 7y = 26 ...

### 4.4 Systems of Equations - Three Variables

It is possible to solve each system several diﬀerent ways. ... − 2x + y − 3z =1 x − 4y+ z =6 4x + 16y+4z = 24 13) ... w + x + y+ z = − 5 3w +2x +2y+4z = − 11

### Solve again pls? 4x-y+2z=-18 -x+2y+z=11 3x+3y-4z=44?

4x − y + 2z = −18 ..... [1] −x + 2y + z = 11 ..... [2] 3x + 3y − 4z = 44 ..... [3] [1] + 4*[2] 7y + 6z = 26 ..... [4] 3*[2] + [3] 9y ...

## Suggested Questions And Answer :

### solve 4x-2y=-2 and 4x+5y=33

line 1: 4x-2y=-2 line 2: 4x+5y=33 line1 - line2: (4x-4x) +(-2y -5y)=-2-33 -7y=-35 y=+5 from line 1, 4x=2y-2  =10-2=8 x=2

### 2x - y = -7 4x+4y=-8

We appear to have two systems of equations, but the method for solving either is essentially the same. First we make the equations into the form y=. The first system: y=2x+7, 4y=-8-4x or y=-2-x.  The second system: y=3-2x, y=2-x. Now we draw the graphs. We can put them all on to the same graph, but we should label all the line graphs so as not to get them mixed up. To draw straight line graphs, we find the x and y intercepts and draw a line joining and passing through them. System 1: y=2x+7: y-int (x=0) is 7, x-int (y=0) is -3.5; y=-2-x: y-int is -2, x-int is -2; system 2: y=3-2x: y-int is 3, x-int is 2; y=2-x: y-int is 2, x-int is 2. For each system, draw a line through the intercepts for each equation. System 1: join 7 on the y-axis to -3.5 on the x axis and join -2 on y to -2 on x; system 2: join 3 on y to 2 on x and join 2 on y to 2 on x. Label the lines with their equations. You should find that in system 1 the lines cross at the solution to 2x+7=-2-x, 3x=-9, so x=-3 and y=1 (-2+3 or -6+7); and in system 2 3-2x=2-x, x=1 and y=1. The points where the lines cross are therefore (-3,1) for system 1 and (1,1) for system 2. These are the graphical representations of the solutions of the two systems of equations.

### consider the given system 4x 4y 4z = 4 12x 13y 14z=15 8x 9y 10z =11 this system is an system and has solucitons. the solution for the system is

Double 1st eqn: 8x+8y+8z=8 and subtract from 3rd: y+2z=3, so y=3-2z. Substitute for y in 2nd and 3rd eqns: 12x+13(3-2z)+14z=15, 12x+39-12z=15, 12(x-z)=-24, x-z=-2, x=z-2. Divide 1st eqn by 4: x+y+z=1 and substitute for x and y: z-2+3-2z+z=1, 1=1, implying many solutions. Check: set z=0, then x=-2, y=3. 1st eqn valid; 2nd eqn valid; 3rd eqn valid. Set z=2, then x=0, y=-1. 1st eqn valid; 2nd eqn valid; 3rd eqn valid. So the system is 3-variable, has many solutions and given z, x=z-2 and y=3-2z.

### word problem please show how

Suppose that in one week, your company takes in \$2,220 in sales at one of its cell phone stores. The price breakdown per phone is as follows: Cell phone model A4: \$35 Smartphone Z20: \$50 Suppose that a total of 51 phones were sold. Set up the system of equations that needs to be solved to determine how many of each type of phone were sold. Give a clear definition of the variables in the system. Solve the system of equations, showing clearly how the solution was determined, and state the results clearly in light of the real-world situation. Verify your results of the 2 linear equations by graphing in the desired graphing program and paste the graph in your assignment document (edit/copy image). Explain how the results are verified by the graph.

### How do I solve equations simultaneously?

In mathematics, simultaneous equations and systems of equations are finite sets of equations whose common solutions are looked for. The systems of equations are usually classified in the same way as the single equations, namely: System of linear equations System of polynomial equations System of ordinary differential equations System of partial differential equations

### Flying bearing 325 degrees 800km,turns course 235 degrees flies 950km - find bearing from A.

1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km. She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown 2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her? There are two coordinate systems used to solve this problem. The first system, used in aeronautics, is a system in which angles are measured clockwise from a line that runs from south to north. The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which runs from left to right. It is necessary to convert the angles stated in the problem to equivalent angles in the rectangular coordinate system. Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the angle we want is the complementary angle measured up from the negative X axis. 90 - 35 = 55 If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A, so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can differentiate between the x and y sides of the triangles by including numbers with the tags. Side y1 is opposite the 55 degree angle, so... y1 / 800 = sin 55 y1 = (sin 55) * 800 y1 = 0.8191 * 800 = 655.32    We'll round that down, to 655 Km Side x1 is opposite the 35 degree angle at the top, so... x1 / 800 = sin 35 x1 = (sin 35) * 800 x1 = 0.5736 * 800 = 458.86     We'll round that up, to 459 Km At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily, we move the origin of the rectangular coordinate system to the point where the turn was made, and proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see that the axis is above the flight path). Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so... y2 / 950 = sin 35 y2 = (sin 35) * 950 y2 = 0.5736 * 950 = 544.89    We'll round that up, to 545 Km Side x2 is opposite this triangle's 55 degree angle, so... x2 / 950 = sin 55 x2 = (sin 55) * 950 x2 = 0.8191 * 950 = 778.19    We'll round that down, to 778 Km The problem asks for the bearing to that second endpoint from the beginning point, which is where we set the first origin. We now draw our third triangle with a hypotenuse from the origin to the second endpoint and its own x and y legs. Because the second flight continued going further out on the negative X axis, we can add the two x values we calculated above. x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km The first leg of the flight was in a northerly direction, but the second leg was in a southerly direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is necessary to subtract the second y value from the first y value to obtain the y coordinate for the triangle we are constructing. y3 = y1 - y2 = 655 Km - 545 Km = 110 Km Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent. tan D = y3 / x3 = 110 / 1237 = 0.0889 Feeding that value into the inverse tangent function, we find the angle that it defines. tan^-1 0.0889 = 5.08 degrees The angle we found is based on the rectangular coordinate system. We need to convert that to the corresponding bearing that was asked for in the problem. We know that the second endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees. We know that the endpoint is above the -X axis, so we must increment the bearing by the size of the angle we calculated. Bearing = 270 + 5.08    approximately 275 degrees The second part of the problem asks how long it will take Wendy (the pilot) to fly straight back to the origin. Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)              = sqrt (2135453) = 1461.319    We'll round that one, too   1461 Km Wendy will fly 1461 Km at 450 Km/hr. How long will that take? t = d / s = 1461Km / (450Km/hr) = 3.25 hours   << 3hrs 15 mins

### The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m.

The temperature in a greenhouse from 7:00p.m. to 7:00a.m. is given by f (t)= 96 - 20sin (t/4), where f (t) is measured in Fahrenheit, and t is the number of hours since 7:00 p.m. A) What is the temperature of the greenhouse at 1:00 a.m. to the nearest Fahrenheit? (B) Find the average temperature between 7:00 p.m. and 7:00 a.m. to the nearest tenth of a degree Fahrenheit. (C) When the temperature of the greenhouse drops below 80 degreeso Fahrenheit, a heating system will automatically be turned on a maintain the temperature at minimum of 80 degrees Fahrenheit. At what values of the to the nearest tenth is the heating system turned on? (D) The cost of heating the greenhouse is \$0.25 per hour for each degree. What is the total cost to the nearest dollar to heat the greenhouse from 7:00 p.m. and 7:00 a.m.?   The equation is: f(t) = 96 – 20sin(t/4), 0 <= t <= 12 (A)  At 1.00 a.m. t = 6 f(6) = 96 – 20.sin(6/4) = 96 – 20*0.99749 = 96 – 19.9499 f(6) = 76 ⁰F (B)  The average temperature would need to be worked out by sampling the temperature at different times throughout the night. Divide the temperature range into N equal intervals, giving N+1 sampling points. We would then have T1 = f(δt), T2 = f(2δt), T3 = f(3δt), ... ,Tn = f(nδt) Where δt = range/N = 12/N, and n = 0..N Giving Tn = 96 – 20.sin((12n/N)/4) = 96 – 20.sin(3n/N) Then Tav = (1/N)*sum(Tn, n = 0 .. N) i.e. Tav = (1/N)*sum(96 – 20.sin(3n/N), n = 0 .. N Tav = 96 – 20. (1/N)*sum(sin(3n/N), n = 0 .. N I used Maple to evaluate the above summation. The results are tabulated as follows.                                  Average Temperature Num Intervals            3       4        6       10       20      50     100    200 Tav over the range 83.39 83.01 82.78 82.69 82.69 82.71 82.72 82.73 As can be seen from the table the temperature is averaging out at:  Tav = 82.7 ⁰F (C)  T = f(t) = 96 – 20sin(t/4), 0 <= t <= 12 At T = 80 ⁰F,            96 – 20sin(t/4) = 80 20.sin(t/4) = 96 – 80 = 16 sin(t/4) = 0.8 t/4 = 0.927295 t = 3.70918 t = 3.7 (to nearest tenth) (D)  The temperature will (normally) drop to 80 ⁰F after t = 3.7 hours and rise again to 80 ⁰F when t = 12 – 3.7 = 8.3 hours. Heating system is turned on for 8.3 – 3.7 = 4.6 hours Cost of heating is 4.6*80*0.25 = 4.6*20 = 92 Cost = \$92

### how do you use brackets in simultaneous equations

One reason you might use brackets in simultaneous equations is for expressing the solution as an ordered set (usually pairs). If, for example, you had a 2-variable system of equations, x and y, you might express the result as the ordered pair (x,y) where x is replaced by the value found for x and y the value found for y. Sometimes a system of equations has more than one solution, so the different solutions would be represented as (x1,y1) and (x2,y2). This can happen when the system involves one or more quadratic equations. The brackets ensure that the values for x and y are not mixed up. Another reason for brackets is when using substitution to solve a system. Suppose there are two equations: ax+by=c and dx+ey=f. From the first equations we can write y=(c-ax)/b and substitute for y in the second equation: dx+e(c-ax)/b=f. That would be the first step in solving. The next step would be to expand the brackets and solve for x in terms of the constants a, b, c, d, e and f. Having found x, you find y by substituting the value of x in y=(c-ax)/b. Sometimes a question involving simultaneous equations comes in a form that uses brackets, so the first step is to expand the brackets, combine like terms, and then continue to solve the system.

### If the solution to a three variable system of linear equations is x+y=3 and z=1, is this a dependent linear system?

There are infinite solutions for x and y but z=1 consistently. There is linear dependence between x and y. z is independent of x and y.