Solve for k 5/8(3k+1)-1/4(7k+2)=1

Mathematics

1 answer:

lesya [120]3 years ago

3 0

Answer:

Step-by-step explanation:

$\dfrac{5}{8}(3k+1)-\dfrac{1}{4}(7k+2)=1\qquad\text{multiply both sides by 8}\\\\8\!\!\!\!\diagup^1\cdot\dfrac{5}{8\!\!\!\!\diagup_1}(3k+1)-8\!\!\!\!\diagup^2\cdot\dfrac{1}{4\!\!\!\!\diagup_1}(7k+2)=8\cdot1\\\\5(3k+1)-2(7k+2)=8\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\(5)(3k)+(5)(1)+(-2)(7k)+(-2)(2)=8\\\\15k+5-14k-4=8\qquad\text{combine like terms}\\\\(15k-14k)+(5-4)=8\\\\k+1=8\qquad\text{subtract 1 from both sides}\\\\k=7$

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