UrbanPro
Login Ask & Answer Success Stories Paid Courses Free Classes Tuition & Classes Fees Write a Review All Categories Help Center
Post your Learning Need Signup as a Tutor
✕

Popular Cities

Bangalore

Hyderabad

Delhi

Mumbai

Chennai

Pune

Kolkata

Gurgaon

Ahmedabad

Jaipur

Noida

✕

Take Class 10 Tuition from the Best Tutors

  • Affordable fees
  • 1-1 or Group class
  • Flexible Timings
  • Verified Tutors

Book a Free Demo

Class 10 > Learn Mathematics > Show that x = 2 and y = 1 satisfy the linear equation...
✕

Search in

Show that x = 2 and y = 1 satisfy the linear equation 2x 3y = 7.

Asked by Last Modified  

31 Answers

Learn Mathematics +1 Class IX-X Tuition

Follow 41
Answer

Please enter your answer

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

2x + 3y=7 Put 2 in place of x and 1 in place of y. 2*2+ 3*1=4+3=7
Comments
Dislike Bookmark

Yes, it satisfy the equation 2x+3y = 7 putting x=2, y= 1, 2(2) + 3(1) = 4 + 3 = 7
Comments
Dislike Bookmark

The equation gets satisfied when we put x=2 and y=1 in 2x + 3y = 7,2(2) + 3(1) = 7.
Comments
Dislike Bookmark
Given, 2x+3y=7. Put the value of x and y in the above equation. 2(2) + 3(1) = 7 4+3=7 7=7
Comments
Dislike Bookmark

Spoken English trainer and french trainer with 4 years of experience.

the equation gets satisfied if u put X=2 and Y=1 for the given equation 2x+3y=7. so 2(2)+3(1)=7 4+3=7 LHS=RHS HENCE SATISFIED
Comments
Dislike Bookmark

Home Tutor

2x + 3y=7. Therefore, put 2 in place of x and 1 in place of y. then the linear equation will be satisfied, 2*2+ 3*1=4+3=7.
Comments
Dislike Bookmark

According to question there is no + sign between 2x 3y. If we assume presences of + sign then it satisfy. Otherwise not.
Comments
Dislike Bookmark

Math Magician

When the given linear equation is written in the form 2x+3y=7 Then we have to substitute the values in the given equation which satisfies it
Comments
Dislike Bookmark

Teacher

On the LHS, if we put the value of x as 2 and y as 1; we will get the equation as 2(2) + 3(1), which after solving comes to 4 + 3, which is equal to RHS i.e. 7. Hence the value of x = 2 and y = 1 satisfy the given equation.
Comments
Dislike Bookmark

Maths Expert

Given, 2x+3y=7. Put the value of x and y in the above equation. 2(2) + 3(1) = 7 4+3=7 7=7
Comments
Dislike Bookmark

View 29 more Answers

Related Questions

PQRS is a square. T and U are the mid-points of sides PS and QR respectively. Find the area of ?OTS, if PQ= 8 cm, where O is the point of intersection of TU and OS.
PQRS IS SQUARE AND T AND U are mid pts of PS &QR, TU//PS AND TU=PS(DINCRVPQRS IS A SQUARE), JOIN QS WHICH INTERSECTS TU AT O, IN ∆PQS ST/TP=1(T IS MID PT OF SP) AND TO//PQ, THEREFORE O IS THE MID POINT...
Aswini
0 0
6

What are the important topics to cover for ICSE Maths for 9th class students. Also please share any sample blueprints for 2019-2020.
Algebra, e.g. quadratic equations, linear equation, geometry triangle, circle, straight line etc
Rajesh Reddy
0 0
5
A triangular park has sides 120 m, 80 m and 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant? Find the cost of fencing it with barbed wire at the rate of Rs. 20 per metre, leaving a space of 3 m wide for a gate on one side.
Perimeter of the garden=120+80+50 =250m As space of 3 m wide for a gate is left on one side, so the length of wire required =(250-3)=247m. Cost of fencing=247X20=Rs 4940. semiperimeter(s)=perimeter/2...
Princy
0 0
5
Write four solutions of 2x 3y = 8.
2x 3y=8 X=1 and y=8/6 X=8/6 and y=1 X=-1 and y=(-8)/6 X=(-8)/6 and y=-1
Gayathri
0 0
6
The sides of a triangular plot are in the ratio of 6 : 7 : 8 and its perimeter is 420 m. Find its area.
8133 metersquare
Pratik

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

Angle ACB = 90 Degree And Angle ACD = x. Find x.
Question: Solution: Method 1 given below: Method 2 given below:

Sujoy Das

2 0
0
many of you have a problem a to square, cube a no..
There is a trick to do it Just remember (a+b)² = a² + 2ab + b² Now remove + sign (a/b)² = a²/2ab/b² You can find square easily Ex will be in a next post Bye
S

Shivam Tiwari

0 0
0
BODMAS Rule
BODMAS stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. In specific regions, PEDMAS (Parentheses, Exponents, Division, Multiplication, Addition and Subtraction) is the synonym...
A

Ashesh Kant Pandey

0 0
0
Solving Linear Wquation in two Varibles By Substitution Method
https://vz-3ad30922-ba4.b-cdn.net/2d534cba-73c9-4441-adee-53ec64e46332/play_480p.mp4

Nikhil Tamoli

1 0
0
Mathemathics Pro Question for 9th to 12th students .
which one is greater ? 6√10 , 3√4, 2√5 Hind to solve this question : take LCM and then find greater Question ask in CGL , CHSL , Bank PO . Question type : Surds and its type .

Savita

0 0
0

Recommended Articles

Meet Sandhya R, a B.Sc tutor from Bangalore

Sandhya is a proactive educationalist. She conducts classes for CBSE, PUC, ICSE, I.B. and IGCSE. Having a 6-year experience in teaching, she connects with her students and provides tutoring as per their understanding. She mentors her students personally and strives them to achieve their goals with ease. Being an enthusiastic...

Read full article >

Meet Swati, a Hindi Tutor from Bangalore

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Read full article >

Quest Academy - Institute of the month

Quest Academy is a professional Bangalore based NEET and JEE (Main + Advanced) training institute. The academy was incorporated in 2015 to cater to the needs of students, who aim to crack competitive exams by connecting with the best brains around. The institute helps students enhance their skills and capabilities through...

Read full article >

Meet Mohammad Wazid, a skilled trainer for...

Mohammad Wazid is a certified professional tutor for class 11 students. He has 6 years of teaching experience which he couples with an energetic attitude and a vision of making any subject easy for the students. Over the years he has developed skills with a capability of understanding the requirements of the students. This...

Read full article >

Find Class 10 Tuition near you
Class 10 Tuition in Delhi
Class 10 Tuition in Bangalore
Class 10 Tuition in Hyderabad
Class 10 Tuition in Kolkata
Class 10 Tuition in Mumbai
Class 10 Tuition in Chennai
Class 10 Tuition in Pune
Class 10 Tuition in Noida
Class 10 Tuition in Lucknow
Class 10 Tuition in Jaipur

Looking for Class 10 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you
X

Looking for Class 10 Tuition Classes?

The best tutors for Class 10 Tuition Classes are on UrbanPro

  • Select the best Tutor
  • Book & Attend a Free Demo
  • Pay and start Learning

Take Class 10 Tuition with the Best Tutors

The best Tutors for Class 10 Tuition Classes are on UrbanPro

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more

ThinkVidya Learning Pvt Ltd © 2010-2024All Rights Reserved