Answer :

trent008

Answer:

(a): x is 3 and ky is -1

(b): k is -2

Step-by-step explanation:

Let: 3x + ky = 8 be equation (a)

x - 2 ky = 5 be equation (b)

Then multiply equation (a) by 2:

→ 6x + 2ky = 16, let it be equation (c)

Then equation (c) + equation (b):

[tex] { \sf{(6 + 1)x + (2 - 2)ky = (16 + 5)}} \\ { \sf{7x = 21}} \\ { \sf{x = 3}}[/tex]

Then ky :

[tex]{ \sf{2ky = 3 - 5}} \\ { \sf{ky = - 1}}[/tex]

[tex]{ \bf{y = \frac{1}{2} }} \\ { \sf{ky = - 1}} \\ { \sf{k = - 2}}[/tex]

MrRoyal

Simultaneous equations are used to represent a system of related equations.

The value of k when [tex]y = \frac 12[/tex] is -2

Given that:

[tex]3x + ky = 8[/tex]

[tex]x - 2ky = 5[/tex]

[tex]y = \frac 12[/tex]

Substitute [tex]y = \frac 12[/tex] in both equations

[tex]3x + ky = 8[/tex]

[tex]3x + k \times \frac 12 = 8[/tex]

[tex]3x + \frac k2 = 8[/tex]

[tex]x - 2ky = 5[/tex]

[tex]x - 2k \times \frac 12 = 5[/tex]

[tex]x - k = 5[/tex]

Make x the subject in [tex]x - k = 5[/tex]

[tex]x = 5 + k[/tex]

Substitute [tex]x = 5 + k[/tex] in [tex]3x + \frac k2 = 8[/tex]

[tex]3(5 + k) + \frac k2 = 8[/tex]

Open bracket

[tex]15 + 3k + \frac k2 = 8[/tex]

Multiply through by 2

[tex]30 + 6k + k = 16[/tex]

[tex]30 + 7k = 16[/tex]

Collect like terms

[tex]7k = 16 - 30[/tex]

[tex]7k = - 14[/tex]

Divide both sides by 7

[tex]k = -2[/tex]

Hence, the value of constant k is -2.

Read more about simultaneous equations at:

https://brainly.com/question/16763389

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