Question

(a) Solve 7( k - 3 ) = 3k - 5

(b) Expand and simplify (2x + 3 )( x - 8)

(c) Solve 7 - 3 f  = 2
                 4

Answer 1

`(a)\\7(k-3)=3k-5\\7(k)+7(-3)=3k-5\\7k-21=3k-5\ \ \ \ |add\ 21\ to\ both\ sides\\7k=3k+16\ \ \ \ |subtract\ 3k\ from\ both\ sides\\4k=16\ \ \ \ \ |divide\ both\ sides\ by\ 4\\\boxed{k=4}`



`(b)\\(2x+3)(x-8)=(2x)(x)+(2x)(-8)+(3)(x)+3(-8)\\\\=2x^2-16x+3x-24=\boxed{2x^2-13x-24}`



`(c)\\(7-3f)/(4)=2\ \ \ \ |multiply\ both\ sides\ by\ 4\\\\\\ot4^1\cdot(7-3f)/(\\ot4_1)=4\cdot2\\\\7-3f=8\ \ \ \ \ |subtract\ 7\ from\ both\ sides\\\\-3f=1\ \ \ \ \ |divide\ both\ sides\ by\ (-3)\\\\\boxed{f=-(1)/(3)}`

Answer 2

a)
`7(k-3)=3k-5\\ 7k-21=3k-5\\ 7k-3k=-5+21\\ 4k=16\\ k=\frac { 16 }{ 4 } \\ k=4`

b)
`(2x+3)(x-8)\\ 2{ x }^( 2 )-16x+3x-24\\ 2{ x }^( 2 )-13x-24`

c)
`\frac { 7-3f }{ 4 } =2\\ 7-3f=4\cdot 2\\ 7-3f=8\\ -3f=8-7\\ -3f=1\\ f=-\frac { 1 }{ 3 }`