Grade 6 / a little further on: algebra 17 exercises corrected

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1. Reverse operations E.2406 1 Complete each of the following diagrams : 2 Using the diagrams above, give the solutions of the fol-lowing equations : a x × 2 = 5 b x + 3 = 7 c x − 4 = 7 d x × 5 = 2 E.10353 On each of the diagrams below, com-plete the missing information : 1 Complete the following diagrams : 2 Give the solutions to each of the following equations : a x + 3 = 5 b 2 × x = 8 c x − 5 = 3 d x + 3.2 = 5 e 6 × x = 9 f x ÷ 2 = 7.5 E.10354 Solve the following equations : a x + 5 = 12 b x − 3 = 2 c 3 × x = 4.2 d x ÷ 5 = 1.2 Hint: one may use a diagram like below to solve these equations : 2. ˇOne-sidedı equations E.10355 Note: for the equation (2 × x )+1=7 , finding the solutions is easy because all you need to do, using the priorities of the operations, to see how the expression on the left-hand side is constructed in order to obtain the numerical value on the right-hand side. The diagram below shows this resolution : The solution to the equation (2 × x )+1=7 is the number 3 . Determine the solution to the equation (7 × x )+3=17 by com-pleting the diagram below : E.2407 1 Complete the lower part of the following diagram: 2 What value of ˇ x ı verifies the equality below : (2 × x ) − 3 = 5 https://chingmath.fr chapExoCorrec/2406 sacados/2406 ×2??5??7 ?÷25??×43? ?5??;42;2? ?−42??÷22? ??10?×36? ×5??0;75?×31? chapExoCorrec/10353 sacados/10353 ??5×2??8−5??3 ;2??5×6??9÷2??7;5 chapExoCorrec/10354 sacados/10354 ???? chapExoCorrec/10355 sacados/10355 x2×x(2×x1763×2÷2−1 x17 chapExoCorrec/2407 sacados/2407 x2×x(2×x−35×2÷2−1
E.10356 Without justification, give the solutions of the following equations : a (3 × x ) + 1 = 4 b (5 × x ) − 4 = 6 c (2 × x ) + 1 = 5 E.10357 Using for each question a diagram sim- ilar to this one above, find the value of ˇ x ı that verifies the equality: a (3 × x ) − 2 = 7 b (5 × x ) + 10 = 16 c (3 × x ) + 1 = 5.5 d 2 × (3 × x ) + 1 = 8 3. ˇTwo-sidedı equations E.10351 Determine, for each question, the value of x achieving equilibrium of the balance : a b c d E.10352 Method : to solve the equation : (7 × x ) + 4 = 2 × x ) + 14 we transform step by step to arrive at the last step in the form ˇ x = : : : ı allowing us to obtain the solution of this equation. Here’s a proposal for a multi-step transformation to obtain the solution of the equation : Solve the following equations : a (5 × x ) + 7 = (3 × x ) + 13 b (2 × x ) + 1 = x + 8 c (9 × x ) + 6 = (5 × x ) + 14 d (4 × x ) + 1 = x + 10 4. Systems of equations: introduction to substitution E.11559 Consider the two purchases below : The price of a can is . . . The price of a coffee is . . . E.11567 We know the following relationships : We have the value : =. . . 5. System of equations: introduction to linear combinations E.11568 We know the following relationships : We have the value : =. . . E.10495 In a café, here are two orders and the amount of their bill: Invoice: 4.5 e . Invoice: 10.8 e In this café, what are the prices of a croissant and a coffee? https://chingmath.fr chapExoCorrec/10356 sacados/10356 chapExoCorrec/10357 sacados/10357 chapExoCorrec/10351 sacados/10351 4x33x5 5x12x10 8x24x14 3x2x6 chapExoCorrec/10352 sacados/10352 (7×x4(2×x14−4−47×x(2×x10−(2×x−(2×x5×x10÷5÷5x2 sacados/11559 ==1124+++10++++ sacados/11567 13610 090180-90 sacados/11568 2823 chapExoCorrec/10495 sacados/10495
E.10497 In a café, here are two orders and the amount of their bill: Invoice A : 15.20 e . Invoice B : 10 e What are the prices, in this café, for a coffee and a can? E.10496 In a café, here are two orders and the amount of their bill: Invoice: 11 e . Invoice: 14.3 e In this café, what are the prices of a croissant and a can? E.11560 Consider the two purchases below : The price of a can is . . . . . . e . The price of a coffee is . . . . . . . E.11561 Consider the two purchases below : The price of a can is . . . . . . e . The price of a coffee is . . . . . . e . https://chingmath.fr chapExoCorrec/10497 sacados/10497 chapExoCorrec/10496 sacados/10496 sacados/11560 ==20e15e++++++++++++ sacados/11561 ==14.5e24e+++++++++++