Answer:
1.) k-5=23
k=23+5
k=28
Checking,
Substitute the values,
28-5=23
23=23
2.) 5x=-35
[tex] \frac{5x}{5} \: = \frac{ - 35}{5} [/tex]
Divide both sides by 5,
x= -7
Checking,
Substitute the values,
5(-7)= -35
-35=-35
3.) 2x-3=7
2x=7+3
2x=10
[tex] \frac{2x}{2} = \frac{10}{2} [/tex]
Divide both sides by 2,
x=5
4.) Equation:
let 5+2x= the second number
let x= the first number
So, the eqution is,
x+(5+2x)=71
x+2x=71-5
3x-66
[tex] \frac{ 3x}{3} = \frac{66}{3} [/tex]
Divide both sides by 3,
x= 22
5.) Equation:
let (25+ n)= students
let n= teachers
So the eqution is,
n+(25+n)=139
n+n=139-25
2n=114
[tex] \frac{2n}{2} = \frac{114}{2} [/tex]
Divide both sides by 2,
n= 57
B.SOLVING LINEAR INEQUALITIES
1.) m -8 ≥ 12
m ≥ 12 + 8
m ≥ 20
2.) 3x + 6 < 18
3x < 18 -6
3x < 12
[tex] \frac{3x}{3} < \frac{12}{3} [/tex]
x< 4
3.) 12 + 3k ≥ -6
3k ≥ -6 - 12
3k ≥ -18
[tex] \frac{3k}{3} \geqslant - \frac{18}{3} [/tex]
Divide both sides by 3,
k ≥ -6
4.) 2-5h < 22
5h < 22-2
5h < 20
[tex] \frac{5h}{5} = \frac{20}{5} [/tex]
Divide both sides by 5,
h= 4
5.) Equation,
let 2n= Ana's age
let n= Lorna's Age
So the Equation is,
2n+n ≤ 72
Now let's solve for their ages
2n+n ≤ 72
3n ≤ 72
[tex] \frac{3n}{3} \leqslant \frac{72}{3} [/tex]
Divide both sides by 3,
n ≤ 24
So,
the age of lorna is n ≤ 24
and the age of Ana is 2n ≤ 24
Step-by-step explanation:
Correct me if I'm wrong.. Pero sanaa po makatulong.
