Mathematics, 28.05.2020 20:58 golderhadashaowtatz
Rahul solved the equation 2(x β ) β 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction right-parenthesis minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction x = 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction right-parenthesis minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction . In which step did he use the addition property of equality? A table titled Rahul's Solution with 2 columns and 5 rows. The first column, Steps, has the entries 1, 2, 3, 4. The second column, Resulting equations, has the entries, 2 x minus StartFraction 1 Over 4 EndFraction minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction, StartFraction 7 Over 5 EndFraction x minus StartFraction 1 Over 4 EndFraction equals StartFraction 55 Over 4 EndFraction, StartFraction 7 Over 5 EndFraction x equals StartFraction 56 Over 4 EndFraction, x equals 10. Step 1 Step 2 Step 3 Step 4
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Rahul solved the equation 2(x β ) β 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction ri...
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