MHB Algebraic equation problem, rectangular flower bed

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To solve the algebraic equations for the rectangular flower bed, two equations are established: x + 10 = 4x - y and x + 20 = 2x + 3y. The goal is to isolate x by first solving for y in the first equation and substituting it into the second equation. Multiple methods, including substitution and elimination, are discussed to find the value of x. Ultimately, the correct approach leads to the conclusion that x can be determined through systematic manipulation of the equations.
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Any ideas on how to begin ?
 

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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?
 
greg1313 said:
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
 
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.
 
greg1313 said:
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
 
x + 20 = 2x + 3(3x - 10)
 
Another method for solving the system for $x$ is elimination. So, you can start with the system Greg posted:

$$x+10=4x-y\tag{1}$$

$$x+20=2x+3y\tag{2}$$

Now, multiply (1) by 3, then add it to (2) thereby eliminating $y$ and obtaining an equation in $x$ alone. :D
 
MarkFL said:
Another method for solving the system for $x$ is elimination. So, you can start with the system Greg posted:

$$x+10=4x-y\tag{1}$$

$$x+20=2x+3y\tag{2}$$

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

$$x+10=4x-y\tag{1}$$
$$x+10-4x=-y\tag{1}$$
$$-3x+10=-y\tag{1}$$

Multiplying this equation by 3

$$-9x+3y=-30\tag{1}$$
$$x+20=2x+3y\tag{2}$$

Moving x to the left hand side of the equation,

$$-x+3y=-20\tag{2}$$

Hope the equations are correct to move on to elimination.
 
mathlearn said:
$$x+10=4x-y\tag{1}$$
$$x+10-4x=-y\tag{1}$$
$$-3x+10=-y\tag{1}$$

Multiplying this equation by 3

$$-9x+3y=-30\tag{1}$$
$$x+20=2x+3y\tag{2}$$

Moving x to the left hand side of the equation,

$$-x+3y=-20\tag{2}$$

Hope the equations are correct to move on to elimination.

No, this is what I had in mind:

Begin with:

$$x+10=4x-y\tag{1}$$

$$x+20=2x+3y\tag{2}$$

Now, multiply (1) by 3:

$$3x+30=12x-3y\tag{1}$$

$$x+20=2x+3y\tag{2}$$

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

$$4x+50=14x$$

Now solve for $x$...:D
 

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