Evaluate the following expressions for a=-3 | b=1/2 | c= 3/5 | d=-4.
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Xy=10
x+y=29
subtract x from both sides for second equation
y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x=
so
x^2+29x+10=0
a=1
b=29
c=10
x=
x=
x=
x=
x=
or 
those are the 2 numbers
aprox=-28.65 and -0.349
x+y=29
subtract x from both sides for second equation
y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x=
so
x^2+29x+10=0
a=1
b=29
c=10
x=
x=
x=
x=
x=
those are the 2 numbers
aprox=-28.65 and -0.349
The cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8
<h3><u>Solution:</u></h3>Let "f" be the cost of 1 fries
Let "h" be the cost of 1 hamburger
<em><u>Given that, tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24</u></em>
8 x cost of 1 hamburger + 5 x cost of 1 fries = 24
8h + 5f = 24 -------- eqn 1
<em><u>The players were still hungry so the coach bought six more hamburgers and two more fries for $16.60</u></em>
6 x cost of 1 hamburger + 2 x cost of 1 fries = 16.60
6h + 2f = 16.60 ------ eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
Multiply eqn 1 by 2
16h + 10f = 48 ------ eqn 3
Multiply eqn 2 by 5
30h + 10f = 83 -------- eqn 4
<em><u>Subtract eqn 3 from eqn 4</u></em>
30h + 10f = 83
16h + 10f = 48
( - ) ----------------------
14h = 35
<h3>h = 2.5</h3>Substitute h = 2.5 in eqn 1
8(2.5) + 5f = 24
20 + 5f = 24
5f = 4
<h3>f = 0.8</h3>Thus cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8