A few algebra problems help pls....?

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in Science & Mathematics

1. Determine the quadrant(s) in which the point is located or the axis on which the point is located without plotting it:

(0, -3)

2. Determine the quadrant(s) in which the point is located or the axis on which the point is located without plotting it:

(x, 5), x < 0

3. Determine the quadrant(s) in which the point is located or the axis on which the point is located without plotting it:

(-6, y), y is a real number

4. Solve the equation for y:

2x + 3y = 6

5. Solve the equation for y:

x - 2y = 8

6. Determine whether the ordered pairs are solutions of the equation:

x - 3y = 4

(a) (1, -1)

(b) (0, 0)

(c) (2, 1)

(d) (5, -2)

7. Determine whether the ordered pairs are solutions of the equation:

y - 2x = -1

(a) (3, 7)

(b) (0, -1)

(c) (-2, -5)

(d) (-1, 0)

Comments

  • fahd-sharifffahd-shariff

    1. y-axis

    2. 2nd quadrant

    3. 2nd or 3rd quadrants

    4. 2x + 3y = 6 => y = 2 - (2/3)x

    5. x - 2y = 8 => y = (1/2)x - 4

    6. x - 3y = 4 => x = 4 + 3y

    Only (a) is a solution

    7. y - 2x = -1 => y = -1 + 2x

    (b), (c) are solutions

  • aaylasecuraaaylasecura

    1. The coordinates for:

    Quadrant 1 are (positive, positive)

    Quadrant 2 are (negative, positive)

    Quadrant 3 are (negative, negative)

    Quadrant 4 are (positive, negative)

    But because (0, -3) is on an axis, it is not in a quadrant. It is on the Y-Axis.

    2. Since x<0, that means it will always be negative. So the (negative, positive) quadrant is Quadrant 2.

    3. Since y can be any real number, but x is negative, it must be a quadrant that has negative x values, in other words, Quadrant 2 and 3.

    4. 2x+3y=6

    Solve for one of the variables and then substitute. I'll solve for y.

    Subtract 2x from both sides: 3y=6-2x

    Divide both sides by 3: y=2-2/3x

    Now subsitute for y: 2x+2-2/3x=6

    Subtract 2 from both sides:2x-2/3x=4

    Combine 2x and -2/3x: 4/3x=4

    Divide by 4/3 (multiply by 3/4): x=3

    5. x-2y=8

    Subtract 2y from both sides:x=8-2y

    Subtitute for x: 8-2y-2y=8

    Combine like terms: 8-4y=8

    Subtract 8 from both sides: -4y=0

    Divide by -4: y=0

    6. Just plug the pairs into the equation to see if they work.

    (a) Yes

    (b)No

    (c) No

    (d)No

    7. Same as problem 6.

    (a) No

    (b) Yes

    (c) Yes

    (d) No

  • ikaika

    1. the point is located on y-axis

    2. quadrant 2

    3. quadrant 2 or 3

    4. 2x + 3y = 6

    3y = -2x +6

    y = -2/3 + 2

    5. x - 2y = 8

    -2y = -x + 8 (divided by -2)

    y = 1/2x - 4

    6. a. (1.-1) because x - 3y = 4 is 1/3x + 4/3

    7. b(0,-1) and c(-2,-5). because y-2x = -1 is equal to y = 2x-1

  • anonymousanonymous

    BTW x^2 skill x to the capability of two or (x-3)^2=255 could be how your first answer is represented on a keyboard enter. and x^(a million/2) is comparable as sq. root X so 255= (x-3)(x-3) prolonged is 255= X^2-6x+9 or 0=X^2-6x-216 sub into Quadratic formula to get x= (6 (plus or minus)(900)^(a million/2))/2 to place that in the time of words x equals 6 plus or minus the sq. root of 900 throughout 2 for this reason x= 18 or -12 Have a circulate at something and notice what you get

  • zachzach

    1) y axis

    2 )4 quadrant

    3) quadrant 3 and 4

    4) -2x+2=y

    5) y=-4+x

    6) (a)

    7) (d)

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