₦airaland Forum

2x-2y=5
2x+3y+1=0

Using graphical method, thanks

2x-2y = 5

x-y = 5/2

X = 5/2+y ------i


2x + 3y = -1

2(5/2+y) +3y = -1

5y = -6

Y= -6/5. --------ii

Sub ii into I

X = 7/5

richmond500:
2x-2y=5
2x+3y+1=0

Using graphical method, thanks

This is simultaneous eqns.

Pm me

ChiefJusticeFuc:
2x-2y = 5

x-y = 5/2

X = 5/2+y ------i


2x + 3y = -1

2(5/2+y) +3y = -1

5y = -6

Y= -6/5. --------ii

Sub ii into I

X = 7/5

thanks boss but I need graphical method

Fidelity2050:
This is simultaneous eqns.

Pm me

pm kwa

richmond500:
pm kwa

Ok na. No worry urself. Thanks

ChiefJusticeFuc:
2x-2y = 5

x-y = 5/2

X = 5/2+y ------i


2x + 3y = -1

2(5/2+y) +3y = -1

5y = -6

Y= -6/5. --------ii

Sub ii into I

X = 7/5

Use graphical method

No graph tools on Nairaland unless one plots the graph elsewhere and export as JPEG here.

Mynd44:
Use graphical method

give me graph

Mynd44:
Use graphical method

that was what I was looking for too, i wasn't looking for elimination method, I can do that one.

Abeg help me move am to front page before my nephew go think say I no sabi book she no go understand say just because I go university no mean say I go fit solve every maths

ChiefJusticeFuc:
give me graph

u fit just write am for book na, Oga sir.
I fit plot graph that year for ordinary paper sef.

Thanks in advance, my pride depends on it

CodeTemplar:
No graph tools on Nairaland unless one plots the graph elsewhere and export as JPEG here.

just rough sketch am, I go fit explain am give the person

2x-2y=5.......(1) x -2
2x+3y+1=0.......(2) x2

Multiply equ 1 by -2
Multiply equ 2 by 2

New equations become:

-4x + 4y = -10.....(3)
4x + 6y = -2........(4)

Adding equ 3 from 4

-4x cancels 4x, 4y + 6y becomes 10y

Then -10 + (-2) becomes -12

We have 10y = -12

y = - 12/10 = -6/5

To get x, from equ 4

4x + 6 y = -2

4x + 6 (-6/5) = -2

4x -(36/5) = -2

4x = -2 + 36/5

4x = (-10+36)/5= 26/5

4x = 26/5

x = 26/5 ÷ 4 = 26/5 x 1/4 = 26/20= 13/10

x = 13/10, y = -6/5

Now, the graph can now be plotted

y = mx + b

For equ 1

x = 5 - 2y /2

For equ 2

y = 2/3 x - 1/3
Plotting the two lines on the coordinates

Equation 1 has a slop of 1 and intersect at y-intercept of -5/2

Equation 2 has a slop of -2/3 and y-intercept of 1/3

Plotting the lines you get ( 13/10, -6/5)

Graph photo coming soon

richmond500:
2x-2y=5
2x+3y+1=0

Using graphical method, thanks

You can't expect someone to do that for you here. Rather, be open to learning and ask them to explain to you.
Take equation 1 and make y the subject of formula
Take equation 2 and make y the subject of formula

Now, select values of x from say -3 down to +3
Set up a first table for equation 1

When x = -3, find y
When x = -2, find y etc etc till you get to the point where X = +3


Set up the second table again with the equation 2 where you made y the subject following similar steps .


You now have 2 tables for 2 different equations

Plot your points on the graph and select any convenient scale of your choice that would cover all values.


Use table 1 to plot the points. When you're done, draw a straight line through the points.

Use table 2 to plot the points there as well. When you're done, draw a straight line through those points.

The two lines are going to intersect. Trace the point of intersection to the y axis and note the point . That's the value of y. Trace the point of intersection to the X axis and note the value. That's the value of x.

That's all!

If you require further explanation, do let me know.

naturalwaves:
You can't expect someone to do that for you here. Rather, be open to learning and ask them to explain to you.
Take equation 1 and make y the subject of formula
Take equation 2 and make y the subject of formula

Now, select values of x from say -3 down to +3
Set up a first table for equation 1

When x = -3, find y
When x = -2, find y etc etc till you get to the point where X = +3


Set up the second table again with the equation 2 where you made y the subject following similar steps .


You now have 2 tables for 2 different equations

Plot your points on the graph and select any convenient scale of your choice that would cover all values.


Use table 1 to plot the points. When you're done, draw a straight line through the points.

Use table 2 to plot the points there as well. When you're done, draw a straight line through those points.

The two lines are going to intersect. Trace the point of intersection to the y axis and note the point . That's the value of y. Trace the point of intersection to the X axis and note the value. That's the value of x.

That's all!

If you require further explanation, do let me know.

person don use mouth explain maths give u before?
This ur explanation sounds like programming language to me..
I appreciate ur effort in helping sha

Use equation 1 and 2 to plot the graph.

richmond500:
thanks boss but I need graphical method

Phabulous4:
Use equation 1 and 2 to plot the graph.

ok

naturalwaves:
You can't expect someone to do that for you here. Rather, be open to learning and ask them to explain to you.
Take equation 1 and make y the subject of formula
Take equation 2 and make y the subject of formula

Now, select values of x from say -3 down to +3
Set up a first table for equation 1

When x = -3, find y
When x = -2, find y etc etc till you get to the point where X = +3


Set up the second table again with the equation 2 where you made y the subject following similar steps .


You now have 2 tables for 2 different equations

Plot your points on the graph and select any convenient scale of your choice that would cover all values.


Use table 1 to plot the points. When you're done, draw a straight line through the points.

Use table 2 to plot the points there as well. When you're done, draw a straight line through those points.

The two lines are going to intersect. Trace the point of intersection to the y axis and note the point . That's the value of y. Trace the point of intersection to the X axis and note the value. That's the value of x.

That's all!

If you require further explanation, do let me know.

I made y subject of formular and I couldn't go any further

Ebuka78:
Graph photo coming soon

I'll be glad bro

To solve the system of equations using the graphical method, we'll plot both equations on the same graph and find their point of intersection. The point of intersection represents the solution to the system.

### Given Equations:
1. \( 2x - 2y = 5 \)
2. \( 2x + 3y + 1 = 0 \)

### Step 1: Rewrite the Equations in Slope-Intercept Form (\( y = mx + b \))
#### Equation 1: \( 2x - 2y = 5 \)
\[
\begin{align*}
2x - 2y &= 5 \\
-2y &= -2x + 5 \\
y &= \frac{-2x + 5}{-2} \\
y &= x - \frac{5}{2}
\end{align*}
\]
So, \( y = x - \frac{5}{2} \).

#### Equation 2: \( 2x + 3y + 1 = 0 \)
\[
\begin{align*}
2x + 3y + 1 &= 0 \\
3y &= -2x - 1 \\
y &= \frac{-2x - 1}{3} \\
y &= -\frac{2}{3}x - \frac{1}{3}
\end{align*}
\]
So, \( y = -\frac{2}{3}x - \frac{1}{3} \).

### Step 2: Plot the Equations
1. **For \( y = x - \frac{5}{2} \):**
- The slope (\( m \)) is 1.
- The y-intercept (\( b \)) is \(-\frac{5}{2}\).

2. **For \( y = -\frac{2}{3}x - \frac{1}{3} \):**
- The slope (\( m \)) is \(-\frac{2}{3}\).
- The y-intercept (\( b \)) is \(-\frac{1}{3}\).

### Step 3: Find the Point of Intersection
The point where the two lines intersect is the solution to the system. From the graph, you can estimate the point of intersection.

### Step 4: Verify the Solution
To ensure accuracy, solve the system algebraically:

#### From Equation 1:
\[
y = x - \frac{5}{2}
\]

#### Substitute \( y \) into Equation 2:
\[
2x + 3\left(x - \frac{5}{2}\right) + 1 = 0
\]
\[
2x + 3x - \frac{15}{2} + 1 = 0
\]
\[
5x - \frac{15}{2} + \frac{2}{2} = 0
\]
\[
5x - \frac{13}{2} = 0
\]
\[
5x = \frac{13}{2}
\]
\[
x = \frac{13}{10}
\]

#### Substitute \( x \) back into Equation 1 to find \( y \):
\[
y = \frac{13}{10} - \frac{5}{2}
\]
\[
y = \frac{13}{10} - \frac{25}{10}
\]
\[
y = -\frac{12}{10}
\]
\[
y = -\frac{6}{5}
Final Answer:
The solution to the system is \( x = \frac{13}{10} \) and \( y = -\frac{6}{5} \). This is the point where the two lines intersect on the graph.

naturalwaves:
You can't expect someone to do that for you here. Rather, be open to learning and ask them to explain to you.
Take equation 1 and make y the subject of formula
Take equation 2 and make y the subject of formula

Now, select values of x from say -3 down to +3
Set up a first table for equation 1

When x = -3, find y
When x = -2, find y etc etc till you get to the point where X = +3


Set up the second table again with the equation 2 where you made y the subject following similar steps .


You now have 2 tables for 2 different equations

Plot your points on the graph and select any convenient scale of your choice that would cover all values.


Use table 1 to plot the points. When you're done, draw a straight line through the points.

Use table 2 to plot the points there as well. When you're done, draw a straight line through those points.

The two lines are going to intersect. Trace the point of intersection to the y axis and note the point . That's the value of y. Trace the point of intersection to the X axis and note the value. That's the value of x.

That's all!

If you require further explanation, do let me know.

I wish I could like this a thousand times! You’ll sure make a good mathematics teacher

richmond500:
2x-2y=5
2x+3y+1=0

Using graphical method, thanks

Equation 1: 2x−2y=5
−2y = −2x+5
 y= x – 5/2  
 y = x − 2.5
find y if x= -3 up to x = 3
x y
-3 -5.5
-2 -4.5
-1 -3.5
0 -2.5
1 -1.5
2 -0.5
3 0.5

Equation 2: 2x + 3y + 1 = 0
3y= −2x − 1  
 y = − 23x −13
y = −2/3x − 1/3
using same -3 to 3 you have the table
x y
-3 1.6667
-2 1
-1 0.3333
0 -0.3333
1 -1
2 -1.6667
3 -2.3333


When you trace the point of intersect, the two lines intersect at (1.3, −1.2), which is the solution

richmond500:
that was what I was looking for too, i wasn't looking for elimination method, I can do that one.

Abeg help me move am to front page before my nephew go think say I no sabi book she no go understand say just because I go university no mean say I go fit solve every maths

You don enter hand today 🤣

SULBELL:
Equation 1: 2x−2y=5
−2y = −2x+5
 y= x – 5/2  
 y = x − 2.5
find y if x= -3 up to x = 3
x y
-3 -5.5
-2 -4.5
-1 -3.5
0 -2.5
1 -1.5
2 -0.5
3 0.5

Equation 2: 2x + 3y + 1 = 0
3y= −2x − 1  
 y = − 23x −13
y = −2/3x − 1/3
using same -3 to 3 you have the table
x y
-3 1.6667
-2 1
-1 0.3333
0 -0.3333
1 -1
2 -1.6667
3 -2.3333


When you trace the point of intersect, the two lines intersect at (1.3, −1.2), which is the solution

Omo, some people sha sabi book.
Thank u

ChiefJusticeFuc:
2x-2y = 5

x-y = 5/2

X = 5/2+y ------i


2x + 3y = -1

2(5/2+y) +3y = -1

5y = -6

Y= -6/5. --------ii

Sub ii into I

X = 7/5

I guess you missed the part where the OP demanded for the problem to be solved using the graphical method.

richmond500:
I made y subject of formular and I couldn't go any further

Very good. You have made a very good progress. Trust me, we will finish this all by yourself.

In the first place you made y the subject where you got : y= x - 5/2, you will substitute different values of x in to it to get y. Let's start from x = -3.
When x = - 3,
y = -3 - 5/2
y= -3 - 2.5
y= - 5.5

When x = -2,
y= -2 -2.5
y= -4.5

Continue like that

Do for all the values of x from -3 to + 3 which are
-3,-2,-1,0,1,2,3.

Do for all these values to get the respective values of y for the first equation.

I could have easily just solved this for you but I want you to learn it. Let me know once you're done with this.

YModulosGodSon:
I wish I could like this a thousand times! You’ll sure make a good mathematics teacher

Thanks bro. I am a teacher

SULBELL:
Equation 1: 2x−2y=5
−2y = −2x+5
 y= x – 5/2  
 y = x − 2.5
find y if x= -3 up to x = 3
x y
-3 -5.5
-2 -4.5
-1 -3.5
0 -2.5
1 -1.5
2 -0.5
3 0.5

Equation 2: 2x + 3y + 1 = 0
3y= −2x − 1  
 y = − 23x −13
y = −2/3x − 1/3
using same -3 to 3 you have the table
x y
-3 1.6667
-2 1
-1 0.3333
0 -0.3333
1 -1
2 -1.6667
3 -2.3333


When you trace the point of intersect, the two lines intersect at (1.3, −1.2), which is the solution

Well done. I wanted him to go through the process of solving by himself , but it is fine. @richmond500, this was what I was trying to explain to you. Do you understand this solution?

Imagine cracking your head for a useless mathematics while old and senile men like tiefnubu are rulling.


To hell with y'all

naturalwaves:
Well done. I wanted him to go through the process of solving by himself , but it is fine. @richmond500, this was what I was trying to explain to you. Do you understand this solution?

I noticed and many others have practically make the process simple for him.
But we have to do our bit.

naturalwaves:
Well done. I wanted him to go through the process of solving by himself , but it is fine. @richmond500, this was what I was trying to explain to you. Do you understand this solution?

thank u bro, but I just wanted to solve it for my sister.
Thanks though I really appreciate ur effort

richmond500:
thank u bro, but I just wanted to solve it for my sister.
Thanks though I really appreciate ur effort

You're welcome