First, I would get h by itself in the first equation. So, subtracting 2x from each side:
H = 20-2X
Then, we would substitute that into the second equation:
X(H+5) = 50
X( 20-2X +5) = 50
Simplifying the left side, we would add 20 and 5
X(25-2X) = 50
Then, distribute the X
25X - 2X^2 = 50
So, we have a quadratic equation. We can get the left side equal to 0. So, subtracting 25X and adding 2X^2 to each side:
0 = 2x^2 - 25x + 50
This factors into:
0 = (2X-5)(X-10)
So, X = 5/2 and 10
So, we would have a couple of solutions.
If X = 5/2, substituting into the first equation:
2(5/2) + H = 20
5 + H = 20
H = 15
If X = 10, then:
2(10) + H = 20
20 + H = 20
H = 0
So, we would have for solutions (X,H):
(5/2, 15) and (10,0)
Good luck, nhl123. I hope this helps.
To solve the system of equations, apply substitution method. To do so, solve for h in the first equation.
Then, substitute this to the second equation.
Set one side equal to zero.
Set each factor equal to zero and solve for x.
`x-10=0` and `2x-5=0`
Next, plug-in the values of x to h=20-2x.
`x=10` , `h=20-2*10=20-20=0`
`x=5/2` , `h=20-2*5/2=20-5=15`
Hence, the solutions to the system of equations are:
> `x=10` and `h=0`
> `x=5/2` and `h=15` .