# 2x+h=20 x(h+5)=50

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`2x+h=20`

`x(h+5)=50`

To solve the system of equations, apply substitution method. To do so, solve for h in the first equation.

`2x+h=20`

`2x-2x+h=20-2x`

`h=20-2x`

Then, substitute this to the second equation.

`x(h+5)=50`

`x(20-2x+5)=50`

`x(-2x+25)=50`

`-2x^2+25x=50`

Set one side equal to zero.

`2x^2-2x^2+25x-25x=2x^2-25x+ 50`

`0=2x^2-25x+50`

Then, factor.

`0=(x-10)(2x-5)`

Set each factor equal to zero and solve for x.

`x-10=0`      and     `2x-5=0`

`x=10`                             `x=5/2`

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`

and,

> `x=5/2` and `h=15` .

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

Hi, nhl123,

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.