Algebraic equation problem, rectangular flower bed

[mathlearn]

View attachment 5947

Any ideas on how to begin ?

 

[Greg]

You can set up two equations in two unknowns:

x + 10 = 4x - y

x + 20 = 2x + 3y

Can you solve this system for x? Do you see why we only need the value of x to find the area of the flower bed?

 

[mathlearn]

You can set up two equations in two unknowns:

x + 10 = 4x - y

x + 20 = 2x + 3y

Can you solve this system for x? Do you see why we only need the value of x to find the area of the flower bed?

Y = 6.25 ,x = 1.58

 

[Greg]

That's incorrect. Solve

x + 10 = 4x - y

for y then substitute that value for y into

x + 20 = 2x + 3y

and solve for x.

Please show your work.

 

[mathlearn]

That's incorrect. Solve

x + 10 = 4x - y

for y then substitute that value for y into

x + 20 = 2x + 3y

and solve for x.

Please show your work.

x + 10 = 4x - y
x + 10 -4x = - y
-3x + 10 = - y
3x - 10 = y

Now substituting that value for y,

x + 20 = 2x + (3x-10)
x + 20 = 5x - 10
-4x=-30
x = 7.5

 

[Greg]

x + 20 = 2x + 3(3x - 10)

 

[MarkFL]

Another method for solving the system for $x$ is elimination. So, you can start with the system Greg posted:

\(\displaystyle x+10=4x-y\tag{1}\)

\(\displaystyle x+20=2x+3y\tag{2}\)

Now, multiply (1) by 3, then add it to (2) thereby eliminating $y$ and obtaining an equation in $x$ alone. :D

 

[mathlearn]

Another method for solving the system for $x$ is elimination. So, you can start with the system Greg posted:

\(\displaystyle x+10=4x-y\tag{1}\)

\(\displaystyle x+20=2x+3y\tag{2}\)

Now, multiply (1) by 3, then add it to (2) thereby eliminating $y$ and obtaining an equation in $x$ alone. :D

\(\displaystyle x+10=4x-y\tag{1}\)
\(\displaystyle x+10-4x=-y\tag{1}\)
\(\displaystyle -3x+10=-y\tag{1}\)

Multiplying this equation by 3

\(\displaystyle -9x+3y=-30\tag{1}\)
\(\displaystyle x+20=2x+3y\tag{2}\)

Moving x to the left hand side of the equation,

\(\displaystyle -x+3y=-20\tag{2}\)

Hope the equations are correct to move on to elimination.

 

[MarkFL]

\(\displaystyle x+10=4x-y\tag{1}\)
\(\displaystyle x+10-4x=-y\tag{1}\)
\(\displaystyle -3x+10=-y\tag{1}\)

Multiplying this equation by 3

\(\displaystyle -9x+3y=-30\tag{1}\)
\(\displaystyle x+20=2x+3y\tag{2}\)

Moving x to the left hand side of the equation,

\(\displaystyle -x+3y=-20\tag{2}\)

Hope the equations are correct to move on to elimination.

No, this is what I had in mind:

Begin with:

\(\displaystyle x+10=4x-y\tag{1}\)

\(\displaystyle x+20=2x+3y\tag{2}\)

Now, multiply (1) by 3:

\(\displaystyle 3x+30=12x-3y\tag{1}\)

\(\displaystyle x+20=2x+3y\tag{2}\)

then add (1) to (2) thereby eliminating $y$ and obtaining an equation in $x$ alone.

\(\displaystyle 4x+50=14x\)

Now solve for $x$...:D