For systems of equations try using graphing, substitution, and elimination. For example
{2x+7y=3}
{x=-4y}
You should first look at if you have a variable that can be substituted (using substitution) and in this case we do! you plug in the x into 2x meaning 2(-4y)+7y=3
1) distribute -8y+7y=3
2) combine like terms in this case -8y+7y= -1y
3) solve -1y=3
y=-3
so currently our solution is (0,-3)
now we solve for x.
we plug our solved variable (y) into 7y
7(-3) and our equation looks like this
2x+7(-3)=3
1) distribute 7(-3)=-21
2) rewrite then solve 2x+(-21)=3
3) isolate variable -21+21 & 3+21
4) 2x=24
5) solve 2/24 = 12
Meaning our solution is (12,-3)
This is how to solve by substitution.
100%/x%=75/15
<span>(100/x)*x=(75/15)*x - </span>we multiply both sides of the equation by x
<span>100=5*x - </span>we divide both sides of the equation by (5) to get x
<span>100/5=x </span>
<span>20=x </span>
<span>x=20</span>
Cupcakes =c
Drinks = d
22 c + 30d = $44.40...eqn 1
30c + 35d = $53.75...eqn 2
Multiply eqn 1 by 7 and eqn 2 by 6
154c +210d =$310.80...eqn 3
180c+ 210d = $322.5...eqn 4
Eqn 4 - eqn 3
26 c = $11.70
C = 11.7/26 = 45c
Subst c = 45c in eqn 1
$9.90 + 30d =$44.40
d = ($44.40-$9.90)/30 = $1.15
