```Question 57442
1)1. Solve the equation 3(2x-1) + x = 5x+3.
{{{highlight(a)}}}. 0
b. 3
c. 3/2
d. no solution
3(2x-1)+x=5x+3
6x-3+x=5x+3
7x-3=5x+3
-5x+7x+3=-5x+5x+3
2x+3=3
2x+3-3=3-3
2x=0
2x/2=0/2
x=0 (a)
:
2)2. Use the quadratic formula to solve {{{y^2+10y+24=0}}}.
a=1, b=10,c=24
{{{highlight(x=(-b+-sqrt(b^2-4ac))/(2a))}}}
{{{x=(-(10)+-sqrt((10)^2-4(1)(24)))/(2(1))}}}
{{{x=(-10+-sqrt(100-96))/(2(1))}}}
{{{x=(-10+-sqrt(4))/2}}}
{{{x=(-10-sqrt(4))/2}}} and {{{x=(-10+sqrt(4))/2}}}
{{{x=(-10-2)/2}}} and {{{x=(-10+2)/2}}}
{{{x=-12/2}}} and {{{x=-8/2}}}
{{{x=-6}}} and {{{x=-4}}}
:
3)3. Solve by factoring: {{{a^2-5a-14 = 0}}}.
a. 5, 14
b. -7, 2
c. 2, 7
{{{highlight(d)}}}. 7, -2
{{{a^2-5a-14=0}}}  Two numbers that multiply to get -14, but add to get -5 are -7 and 2.  -7*2=-14 and -7+2=-5
{{{(a-7)(a+2)=0}}}
{{{a-7=0}}} and {{{a+2=0}}}
{{{a=7}}} and {{{a=-2}}}
:
4)4. Solve the inequality 3x + 5< with - under < 7  . Write the solution using interval notation.
a. (- infinity; , 4]
b. [4, infinity )
{{{highlight(c)}}}. (-infinity, 2/3]
d. [2/3, infinity )
{{{3x+5<=7}}}
{{{3x+5-5<=7-5}}}
{{{3x<=2}}}
{{{3x/3<=2/3}}}
{{{x<=2/3}}}
(-infinity,2/3]
Happy Calculating!!!```