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Click here 👆 to get an answer to your question ️ 15-5(4c-7)=50
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To solve a linear equation in one variable using balancing method we need to first perform sign rule(s) of multiplication if there is/are bracket(s) present in the given input. Now we need to get rid of constant term present in the LHS and for that we need to add or subtract constant both sides of the equation and perform simplification. After that we need to get rid of the variable term from.
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Solve for c 15-5 (4c-7)=50. 15 − 5 (4c − 7) = 50 15 - 5 ( 4 c - 7) = 50. Simplify 15 −5(4c−7) 15 - 5 ( 4 c - 7). Tap for more steps. −20c+ 50 = 50 - 20 c + 50 = 50. Move all terms not containing c c to the right side of the equation. Tap for more steps. −20c = 0 - 20 c = 0. Divide each term in −20c = 0 - 20 c = 0 by −20 - 20.
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Click here 👆 to get an answer to your question ️ 15-5(4c-7)=50.. Explanation: Reduce the greatest common factor on both sides of the equation: 3-(4c-7)=10 Remove the parentheses: 3-4c-7=10 Rearrange unknown terms to the left side of the equation: -4c=10-3-7-4c=10-3-7 Calculate the sum or difference: -4c=0-4c=0-4c=0 Divide both sides of.
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Solve for x 15-5 (4x-7)=50. 15 − 5 (4x − 7) = 50 15 - 5 ( 4 x - 7) = 50. Simplify 15 −5(4x−7) 15 - 5 ( 4 x - 7). Tap for more steps. −20x+ 50 = 50 - 20 x + 50 = 50. Move all terms not containing x x to the right side of the equation. Tap for more steps. −20x = 0 - 20 x = 0. Divide each term in −20x = 0 - 20 x = 0 by −20 - 20.
1. 4(7a+5)=160 2. 3k+2(5k3)=7 3. 79=7w+3(4w1) 4. 94(2p1)=41 5. y
Solve linear equation with one unknown 15-5(4c-7)=50: Tiger Algebra not only solves linear equations with one unknown 15-5(4c-7)=50, but its clear, step-by-step explanation of the solution helps to better understand and remember the method
