Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, set x3 = t and solve for x1 and x2.) x1 − 3x3 = −7 3x1 + x2 − 2x3 = −5 2x1 + 2x2 + x3 = −3

x1 = -7+3t

x2 = (11-7t)/2

Step-by-step explanation:

Given the system of equations

x1 − 3x3 = −7 ..., 1

3x1 + x2 − 2x3 = −5 2

2x1 + 2x2 + x3 = −3 3

Setting x3 = t.

Substitute x3 = t into equation 1 to get x1 in terms of t.

x1 − 3x3 = −7

x1-3(t) = -7

x1-3t = -7

Add 3t to both sides;

x1-3t+3t = -7+3t

x1 = -7+3t

To get x2, substitute x1 = -7+3t and x3 = t into equation 3

2x1 + 2x2 + x3 = −3

2(-7+3t)+2x2+t = -3

Open the parenthesis

-14+6t+2x2+t = -3

Collect like terms

2x2+6t+t = -3+14

2x2+7t = 11

Subtract 7t from both sides

2x2+7t-7t = 11-7t

2x2= 11-7t

Divide both sides by 2

2x2/2 = (11-7t)/2

x2 = (11-7t)/2

Hence the solution to the system if equation (x1, x2, x3) = (-7+3t, (11-7t)/2, t)

i think it is 18

step-by-step explanation:

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.