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Solve for x and y, if: a) 3x + 4y = -5 and 5x + 6y = -7 c) x + 4y = 3 and 2x + y = -1 e) 2x+y−3=0 and y = 9 + 4x g) 2x−3y−3= 0 and x + y − 4 = 0

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8 parts
a

x=1,y=−2x=1, y=-2

b

x=3,y=2x=3, y=2

c

x=−1,y=1x=-1, y=1

d

x=3,y=2x=3, y=2

e

x=−1,y=5x=-1, y=5

f

x=3,y=1x=3, y=1

g

x=3,y=1x=3, y=1

h

x=14,y=252x=14, y=\frac{25}{2}

Step-by-step solution

26 steps

Here are the solutions for each system of equations:

Part a· 3 steps

We have the system:

  1. 3x+4y=−53x + 4y = -5
  2. 5x+6y=−75x + 6y = -7

  1. Multiply equation (1) by 3 and equation (2) by 2 to make the coefficients of yy equal.

    3(3x+4y)=3(−5)  ⟹  9x+12y=−15(3)3(3x + 4y) = 3(-5) \implies 9x + 12y = -15 \quad (3) 2(5x+6y)=2(−7)  ⟹  10x+12y=−14(4)2(5x + 6y) = 2(-7) \implies 10x + 12y = -14 \quad (4)

  2. Subtract equation (3) from equation (4) to eliminate yy.

    (10x+12y)−(9x+12y)=−14−(−15)(10x + 12y) - (9x + 12y) = -14 - (-15) 10x−9x=−14+1510x - 9x = -14 + 15 x=1x = 1

  3. Substitute x=1x=1 into equation (1).

    3(1)+4y=−53(1) + 4y = -5 3+4y=−53 + 4y = -5 4y=−5−34y = -5 - 3 4y=−84y = -8 y=−2y = -2

    The solution is x=1,y=−2\boxed{x=1, y=-2}.

Part b

x=3,y=2x=3, y=2

We have the system:

  1. 3x−4y=13x - 4y = 1
  2. 7x+y=237x + y = 23

  1. From equation (2), express yy in terms of xx.

    y=23−7x(3)y = 23 - 7x \quad (3)

  2. Substitute equation (3) into equation (1).

    3x−4(23−7x)=13x - 4(23 - 7x) = 1 3x−92+28x=13x - 92 + 28x = 1 31x−92=131x - 92 = 1 31x=9331x = 93 x=9331x = \frac{93}{31} x=3x = 3

  3. Substitute x=3x=3 into equation (3).

    y=23−7(3)y = 23 - 7(3) y=23−21y = 23 - 21 y=2y = 2

    The solution is x=3,y=2\boxed{x=3, y=2}.

Part c

x=−1,y=1x=-1, y=1

We have the system:

  1. x+4y=3x + 4y = 3
  2. 2x+y=−12x + y = -1

  1. From equation (1), express xx in terms of yy.

    x=3−4y(3)x = 3 - 4y \quad (3)

  2. Substitute equation (3) into equation (2).

    2(3−4y)+y=−12(3 - 4y) + y = -1 6−8y+y=−16 - 8y + y = -1 6−7y=−16 - 7y = -1 −7y=−1−6-7y = -1 - 6 −7y=−7-7y = -7 y=1y = 1

  3. Substitute y=1y=1 into equation (3).

    x=3−4(1)x = 3 - 4(1) x=3−4x = 3 - 4 x=−1x = -1

    The solution is x=−1,y=1\boxed{x=-1, y=1}.

Part d

x=3,y=2x=3, y=2

We have the system:

  1. 2x−8=−y  ⟹  2x+y=82x - 8 = -y \implies 2x + y = 8
  2. 7x+4y=297x + 4y = 29

  1. From the rearranged equation (1), express yy in terms of xx.

    y=8−2x(3)y = 8 - 2x \quad (3)

  2. Substitute equation (3) into equation (2).

    7x+4(8−2x)=297x + 4(8 - 2x) = 29 7x+32−8x=297x + 32 - 8x = 29 −x+32=29-x + 32 = 29 −x=29−32-x = 29 - 32 −x=−3-x = -3 x=3x = 3

  3. Substitute x=3x=3 into equation (3).

    y=8−2(3)y = 8 - 2(3) y=8−6y = 8 - 6 y=2y = 2

    The solution is x=3,y=2\boxed{x=3, y=2}.

Part e

x=−1,y=5x=-1, y=5

We have the system:

  1. 2x+y−3=0  ⟹  2x+y=32x + y - 3 = 0 \implies 2x + y = 3
  2. y=9+4xy = 9 + 4x

  1. Substitute equation (2) directly into the rearranged equation (1).

    2x+(9+4x)=32x + (9 + 4x) = 3 6x+9=36x + 9 = 3

  2. Solve for xx.

    6x=3−96x = 3 - 9 6x=−66x = -6 x=−1x = -1

  3. Substitute x=−1x=-1 into equation (2).

    y=9+4(−1)y = 9 + 4(-1) y=9−4y = 9 - 4 y=5y = 5

    The solution is x=−1,y=5\boxed{x=-1, y=5}.

Part f

x=3,y=1x=3, y=1

We have the system:

  1. x+2y−5=0  ⟹  x+2y=5x + 2y - 5 = 0 \implies x + 2y = 5
  2. 6x+11y=296x + 11y = 29

  1. From the rearranged equation (1), express xx in terms of yy.

    x=5−2y(3)x = 5 - 2y \quad (3)

  2. Substitute equation (3) into equation (2).

    6(5−2y)+11y=296(5 - 2y) + 11y = 29 30−12y+11y=2930 - 12y + 11y = 29 30−y=2930 - y = 29

  3. Solve for yy.

    −y=29−30-y = 29 - 30 −y=−1-y = -1 y=1y = 1

  4. Substitute y=1y=1 into equation (3).

    x=5−2(1)x = 5 - 2(1) x=5−2x = 5 - 2 x=3x = 3

    The solution is x=3,y=1\boxed{x=3, y=1}.

Part g

x=3,y=1x=3, y=1

We have the system:

  1. 2x−3y−3=0  ⟹  2x−3y=32x - 3y - 3 = 0 \implies 2x - 3y = 3
  2. x+y−4=0  ⟹  x+y=4x + y - 4 = 0 \implies x + y = 4

  1. From the rearranged equation (2), express xx in terms of yy.

    x=4−y(3)x = 4 - y \quad (3)

  2. Substitute equation (3) into the rearranged equation (1).

    2(4−y)−3y=32(4 - y) - 3y = 3 8−2y−3y=38 - 2y - 3y = 3 8−5y=38 - 5y = 3

  3. Solve for yy.

    −5y=3−8-5y = 3 - 8 −5y=−5-5y = -5 y=1y = 1

  4. Substitute y=1y=1 into equation (3).

    x=4−1x = 4 - 1 x=3x = 3

    The solution is x=3,y=1\boxed{x=3, y=1}.

Part h

x=14,y=252x=14, y=\frac{25}{2}

We have the system (assuming "2x2y3 = 0" means 2x−2y−3=02x - 2y - 3 = 0):

  1. 2x−2y−3=0  ⟹  2x−2y=32x - 2y - 3 = 0 \implies 2x - 2y = 3
  2. 8y=7x+2  ⟹  7x−8y=−28y = 7x + 2 \implies 7x - 8y = -2

  1. Multiply equation (1) by 4 to make the coefficients of yy equal.

    4(2x−2y)=4(3)  ⟹  8x−8y=12(3)4(2x - 2y) = 4(3) \implies 8x - 8y = 12 \quad (3)

  2. Subtract equation (2) from equation (3) to eliminate yy.

    (8x−8y)−(7x−8y)=12−(−2)(8x - 8y) - (7x - 8y) = 12 - (-2) 8x−7x=12+28x - 7x = 12 + 2 x=14x = 14

  3. Substitute x=14x=14 into equation (1).

    2(14)−2y=32(14) - 2y = 3 28−2y=328 - 2y = 3 −2y=3−28-2y = 3 - 28 −2y=−25-2y = -25 y=−25−2y = \frac{-25}{-2} y=252y = \frac{25}{2}

    The solution is x=14,y=252\boxed{x=14, y=\frac{25}{2}}.

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