adriraven123
contestada

Solve using any appropriate method. If the system has an infinite number of solutions, use set-builder notation to write
the solution set. If the system has no solution, state this
13) 4x + 8y = 8
5x + 7y = 1
A) (-8,6)
B) (4,3)
C) (-4,3)
D) No solution
14) 7x + 7y = 8
5x + 5y = 10
A) (10,5)
D) (1,1)
C) {(x,y)|7x + 7y = 8)
B) No solution
15) y = 3x + 4
5x + y = -92
A) No solution
C) (4,8)
B) ((x, y) 5x + y = -92)
D) (-4,-8)
16) 0.4x + 0.4y = 4
0.4x - 0.4y = 1
D)
B)
(659
0(*)
C) (1.4)
D) {x,y) x = 7+ 4y}
17) x - 4y = 1
x= 7+ 4y
A) No solution
B) {(x,y)|x - 4y = 1)​

Respuesta :

MrRoyal

Answer:

[tex]13.\ (x,y) = (-4,3)[/tex]

[tex]14.\ No\ Solution[/tex]

[tex]15.\ (x,y) = (-12,-32)[/tex]

[tex]16.\ (x,y) = (3.75,6.25)[/tex]

[tex]17.\ No\ Solution[/tex]

Explanation:

13)

[tex]4x + 8y = 8[/tex] ---- (1)

[tex]5x + 7y = 1[/tex] ------ (2)

Multiply equation by 5 and equation 2 by 4

[tex]5(4x + 8y = 8)[/tex]

[tex]4(5x + 7y = 1)[/tex]

[tex]5 * 4x + 5 * 8y = 5 * 8\\4*5x + 4*7y = 4 *1[/tex]

[tex]20x + 40y = 40[/tex] ------ (3)

[tex]20x + 28y = 4[/tex] ------ (4)

Subtract equation 3 from 4

[tex](20x + 28y = 4) - (20x + 40y = 40)[/tex]

[tex]20x - 20x+ 28y -40y= 4-40[/tex]

[tex]-12y = -36[/tex]

Divide both sides by -12

[tex]\frac{-12y}{-12} = \frac{-36}{-12}[/tex]

[tex]y = 3[/tex]

Substitute [tex]y = 3[/tex] in equation 1

[tex]4x + 8y = 8[/tex]

[tex]4x + 8(3) = 8[/tex]

[tex]4x + 24 = 8[/tex]

Subtract 24 from both sides

[tex]4x + 24 - 24 = 8 - 24[/tex]

[tex]4x = -16[/tex]

Divide both sides by 4

[tex]\frac{4x}{4} = \frac{-16}{4}[/tex]

[tex]x = -4[/tex]

[tex]Hence\ (x,y) = (-4,3)[/tex]

----------------------------------------------

14)

[tex]7x + 7y = 8[/tex] ---- (1)

[tex]5x + 5y = 10[/tex] ----- (2)

Make x the subject of formula in (2)

[tex]5x + 5y = 10[/tex]

[tex]5x + 5y - 5y = 10 - 5y[/tex]

[tex]5x = 10 - 5y[/tex]

[tex]\frac{5x}{5} = \frac{10 - 5y}{5}[/tex]

[tex]x = \frac{10}{5} - \frac{5y}{5}[/tex]

[tex]x = 2 - y[/tex]

Substitute [tex]x = 2 - y[/tex] in (1)

[tex]7x + 7y = 8[/tex]

[tex]7(2-y) + 7y = 8[/tex]

Open bracket

[tex]7*2-7*y + 7y = 8[/tex]

[tex]14 -7y + 7y = 8[/tex]

[tex]14 \neq 8[/tex]

At this point, we conclude that thee system has no solution.

----------------------------------------------

15)

[tex]y = 3x + 4[/tex] --- 1

[tex]5x + y = -92[/tex] ----2

Substitute [tex]y = 3x + 4[/tex] in (2)

[tex]5x + 3x + 4 = -92[/tex]

[tex]8x + 4 = -92[/tex]

Subtract 4 from both sides

[tex]8x + 4 -4= -92 -4[/tex]

[tex]8x = -96[/tex]

Divide both sides by 8

[tex]\frac{8x}{8} = \frac{-96}{8}[/tex]

[tex]x = -12[/tex]

Substitute [tex]x = -12[/tex] in equation 1

[tex]y = 3x + 4[/tex]

[tex]y = 3(-12) + 4[/tex]

[tex]y = -36 + 4[/tex]

[tex]y = -32[/tex]

[tex]Hence\ (x,y) = (-12,-32)[/tex]

----------------------------------------------

16)

[tex]0.4x + 0.4y = 4[/tex]

[tex]0.4x - 0.4y = 1[/tex]

Add both equations

[tex](0.4x + 0.4y = 4) + (0.4x - 0.4y = 1)[/tex]

[tex]0.4x+0.4x + 0.4y-0.4y = 4+1[/tex]

[tex]0.8x = 5[/tex]

Divide both sides by 0.8

[tex]\frac{0.8x}{0.8} = \frac{5}{0.8}[/tex]

[tex]x = 6.25[/tex]

Substitute [tex]x = 6.25[/tex] in (1)

[tex]0.4(6.25) + 0.4y = 4[/tex]

[tex]2.5 + 0.4y = 4[/tex]

Subtract 2.5 from both sides

[tex]2.5 -2.5 + 0.4y = 4 -2.5[/tex]

[tex]0.4y = 1.5[/tex]

Divide both sides by 0.4

[tex]\frac{0.4y}{0.4} = \frac{1.5}{0.4}[/tex]

[tex]y = \frac{1.5}{0.4}[/tex]

[tex]y = 3.75[/tex]

[tex]Hence,\ (x,y) = (3.75,6.25)[/tex]

----------------------------------------------

17)

[tex]x - 4y = 1[/tex]

[tex]x= 7+ 4y[/tex]

Substitute [tex]x= 7+ 4y[/tex] in (1)

[tex]7 + 4y - 4y = 1[/tex]

[tex]7 \neq 1[/tex]

At this point, we conclude that thee system has no solution.