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Q: What are the intercepts of 2x + 3y – 6z = 30?

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The intercepts of 2x + 3y - 6z = 30 are(15, 0, 0), (0, 10, 0) and (0, 0 , -5).

x: x = 15 y: y = 10 z: z = -5

The result will be a plane that intercepts the x-, y-, and z-axes at +9, +6, and +3, respectively.

The list of choices that accompanies the question is really short.2x + 3y + 6z = 54At the y-intercept, 'x' and 'z' are zero.3y = 54y = 18

Since there are no given triplets, none of them!

6z =

6z-513 = -507

(x + 6z)(x - 6z)

-5x+7y-6z

6z + 28 = 12z + 10 18 = 6z 3 = z

you treat the letters like numbers i.e if x=4y, rearrange for y solution: divide both sides by 4 therefore y=1/4 x example 2: 6z+x/y = 2x-y rearrange for z solution: multiply both sides by y: 6z+x = y(2x-y) 6z+x = 2xy-y2 Take away x from both sides: 6z = 2xy-y2-x divide both sides by 6: z= 1/6(2xy-y2-x) done!

2

6z+28=12z+10 6z-12z=10-28 addition property of equality. -6z=-18 simplification z=3 multiplication property of equality.

14

If 6z=17, divide both sides by 6 z=17/6

-6z + 16 = 52 subtract 16 from each side -6z = 36 divide by -6 z = -6

6z - 8 - 2 + 3 = 0 combining all the constant terms: 6z -7 = 0 adding 7 to both sides: 6z = 7 dividing both sides by 6: z = 7/6

17z + 34 = 6z + 5 17z - 6z + 34 = 6z - 6z + 5 11z + 34 = 5 11z + 34 - 34 = 5 - 34 11z = -29 11z/11 = -29/11 z = -2 7/11 or -29/11 Double check your answer: 17z + 34 = 6z + 5 17(-2 7/11) + 34 = 6z + 5 -44 9/11 + 34 = 6z + 5 -10 9/11 = 6z + 5 -10 9/11 = 6(-2 7/11) + 5 -10 9/11 = -15 9/11 + 5 -10 9/11 = -10 9/11 17z + 34 = 6z + 5 when Z = -2 7/11

-216z3

4x + 3y + 2z = 34; 2x + 4y + 3z = 45; 3x + 2y + 4z = 47 First, eliminate terms in z from 2 of the equations, by muliplying first equation by 2 and subtracting third equation from the answer: 8x + 6y + 4z = 68, subtract leaving 5x + 4y = 21 (equation 4) Similarly multiply the first equation by 3 and the second by 2 giving 12x + 9y + 6z = 102 and 4x + 8y + 6z = 90 Subtract again and we have 8x + y = 12 or y = 12 - 8x Substitute this in equation 4 gives 5x + 4(12 -8x) = 21 Simplify: 5x + 48 - 32x = 21 = -27x = -27 so x = 1 y = 12 - 8x so y = 4 and in one of the original equations 4 + 12 + 2z = 34, ie 2z = 34 -16 so z =9 Check: 2x + 4y + 3z = 2 +16 + 27 = 45 and 3x + 2y + 4z = 3 + 8 + 36 = 47 QED!

15z+5+6z=0 21z+5=0 z=-5/21

6z-4z-z = 6-7-4 z = -5

-6z + 2 = 3z + 4z + 28 Add 6z to both sides: 2 = 3z + 4z + 28 + 6z = 28 + 13z Subtract 28 from both sides: - 26 = 13z Divide both sides by 13: -2 = z

If you mean: 2x-5y+2z = 16 and 3x+2y-3z = -19 and 4x-3y+4z = 18 then the solutions are found as follows:- 3(2x-5y+2z = 16) => 6x-15y+6z = 48 2(3x+2y-3z = -19) => 6x+4y-6z = -38 Adding the above: 12x-11y = 10 thus eliminating z 4(3x+2y-3z = -19) => 12x+8y-12z = -76 3(4x-3y+4z = 18) => 12x-9y+12z = 54 Adding the above: 24x-y = -22 thus eliminating z 2(12x-11y = 10) => 24x-22y = 20 1(24x-y = -22) => 24x-y = -22 Subtracting the above: -21y = 42 thus eliminating x If: -21y = 42 then y = -2 So by substitution: x = -1, y = -2 and z = 4 Check: (2*-1)-(5*-2)+(2*4) = 16 Check: (3*-1)+(2*-2)-(3*4) = -19 Check: (4*-1)-(3*-2)+(4*4) = 18

z equals 7

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