Answer:
[tex]\large\boxed{b.\ x=\dfrac{14}{27}\ \text{and}\ c.\ x=2}[/tex]
Step-by-step explanation:
[tex]4|5-5x|=7x+6\\4|-5(x-1)|=7x+6\\4|-5||x-1|=7x+6\\(4)(5)|x-1|=7x+6\\20|x-1|=7x+6\\\\\text{First step:}\\\text{Based on the de}\text{finition of the absolute value}\\\\|x-1|=\left\{\begin{array}{ccc}x-1&\text{for}\ x\geq1\\1-x&\text{for}\ x<1\end{array}\right\\\\\text{Let}\ x<1\to x\in(-\infty,\ 1).\ \text{Then}\ |x-1|=1-x:\\\\20(1-x)=7x+6\qquad\text{use the distributive property}\\20-20x=7x+6\qquad\text{subtract 20 from both sides}\\-20x=7x-14\qquad\text{subtract}\ 7x\ \text{from both sides}\\-27x=-14\qquad\text{divide both sides by (-27)}\\x=\dfrac{14}{27}<1\qquad \bold{:)}[/tex]
[tex]\text{Let}\ x\geq0\to x\in\left<1,\ \infty\right).\ \text{Then}\ |x-1|=x-1:\\\\20(x-1)=7x+6\qquad\text{use the distributive property}\\20x-20=7x+6\qquad\text{add 20 to both sides}\\20x=7x+26\qquad\text{subtract}\ 7x\ \text{from both sides}\\13x=26\qquad\text{divide both sides by 13}\\x=2\geq1\qquad \bold{:)}[/tex]
