Question

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Please solve this problem with Step-by-step explanation


a table cost ₹ 40 more than a chair. the total cost of 3 tables and 2 chairs is Rs 745 find the cost of each. ​

Answer 1

Let the cost of table be ₹x

Let the cost of table be ₹xLet the cost of chair be ₹y

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given condition

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*3

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:Replace value of y in equation (1)

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:Replace value of y in equation (1)x-125=40

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:Replace value of y in equation (1)x-125=40x=40+125

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:Replace value of y in equation (1)x-125=40x=40+125x=165

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:Replace value of y in equation (1)x-125=40x=40+125x=165Therefore,

Let the cost of table be ₹xLet the cost of chair be ₹yAccording to the given conditionx-y=40.......(1)3x+2y=745........(2)Multiplying equation (1) by 3(x-y=40)*33x-3y=120.........(3)Subtracting eqaution (2) and (3) we get5y=625y=125Solving for x:Replace value of y in equation (1)x-125=40x=40+125x=165Therefore,Cost of table = ₹x = ₹165

₹165Cost of chair = ₹y = ₹125

Answer 2

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