Question

2 – Calcule a raiz ou solução das seguintes equações:
Com soma por favor

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm Grupo~1}\\\tt a)~\sf 5x+1=36\\\sf 5x=36-1\\\sf 5x=35\\\sf x=\dfrac{35}{5}\\\\\sf x=7\\\tt b)~\sf 7x=4x+5\\\sf 7x-4x=5\\\sf 3x=5\\\sf x=\dfrac{5}{3}\\\tt c)~\sf 9x-7=5x+13\\\sf 9x-5x=13+7\\\sf 4x=20\\\sf x=\dfrac{20}{4}\\\\\sf x=5\\\tt d)~\sf15x-13=-39-11x\\\sf 15x+11x=13-39\\\sf 26x=-26\\\sf x=-\dfrac{26}{26}\\\\\sf x=-1\\\tt e)~\sf21+9x=15x+33\\\sf 9x-15x=33-21\\\sf -6x=12\cdot(-1)\\\sf 6x=-12\\\sf x=-\dfrac{12}{6}\\\\\sf x=-2\end{array}}$

$\Large\boxed{\begin{array}{l}\tt f)~\sf16-11x+12=-15x\\\sf -11x+15x=-12-16\\\sf 4x=-28\\\sf x=-\dfrac{28}{4}\\\\\sf x=-7\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Grupo~2}\\\tt a)~\sf2\cdot(2x-1)-6\cdot(1-2x)=2\cdot(4x-5)\\\sf 4x-2-6+12x=8x-10\\\sf 4x+12x-8x=-10+2+6\\\sf4x=-2\\\sf x=-\dfrac{2\div2}{4\div2}\\\\\sf x=-\dfrac{1}{2}\\\\\tt b)~\sf 2x+3\cdot(x-2)=7x+34\\\sf 2x+3x-6=7x+34\\\sf 2x+3x-7x=34+6\\\sf-2x=40\cdot(-1)\\\sf 2x=-40\\\sf x=-\dfrac{40}{2}\\\\\sf x=-20\\\tt c)~\sf4\cdot(x-2)=4+2\cdot(x-1)\\\sf 4x-8=4+2x-2\\\sf 4x-2x=4+8-2\\\sf 2x=10\\\sf x=\dfrac{10}{2}\\\\\sf x=5\end{array}}$

$\boxed{\begin{array}{l}\tt d)~\sf10\cdot(100-x)+2x=3\cdot(x+9)-28\\\sf1000-10x+2x=3x+27-28\\\sf -10x+2x-3x=27-28-1000\\\sf-11x=-1001\cdot(-1)\\\sf 11x=1001\\\sf x=\dfrac{1001}{11}\\\\\sf x=91\\\tt e)~\sf 5\cdot(x+60)-400,7=8\cdot(x-40)\\\sf 5x+300-400,7=8x-320\\\sf 5x-8x=320-300+400,7\\\sf-3x=420,7\cdot(-1)\\\sf 3x=-420,7\\\sf x=-\dfrac{420,7\cdot10}{3\cdot10}\\\\\sf x=-\dfrac{4207}{30}\end{array}}$

$\large\boxed{\begin{array}{l}\tt f)~\sf7\cdot(2x-50)-4x=10\cdot(51,9-0,1x)\\\sf 14x-350-4x=519-x\\\sf 14x-4x+x=519+350\\\sf11x=869\\\sf x=\dfrac{869}{11}\\\\\sf x=79\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Grupo~3}\\\tt a)~\sf\dfrac{3x}{4}-\dfrac{2}{3}=x-\dfrac{5}{2}\cdot(12)\\\\\sf9x-8=12x-30\\\sf 9x-12x=8-30\\\sf -3x=-24\cdot(-1)\\\sf 3x=24\\\sf x=\dfrac{24}{3}\\\\\sf x=8\\\tt b)~\sf\dfrac{x}{2}+1=\dfrac{x}{5}+\dfrac{1}{4}\cdot(20)\\\\\sf 10x+20=4x+5\\\sf 10x-4x=5-20\\\sf 6x=-15\\\sf x=-\dfrac{15\div3}{6\div3}\\\\\sf x=-\dfrac{5}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{x}{4}+\dfrac{x}{3}=x-100\cdot(12)\\\\\sf3x+4x=12x-1200\\\sf 3x+4x-12x=-1200\\\sf-5x=-1200\cdot(-1)\\\sf 5x=1200\\\sf x=\dfrac{1200}{5}\\\\\sf x=240\\\tt d)~\sf\dfrac{2x}{3}+\dfrac{5x}{6}=\dfrac{1}{2}\cdot(6)\\\sf 4x+5x=3\\\sf 9x=3\\\sf x=\dfrac{3\div3}{9\div3}\\\\\sf x=\dfrac{1}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\tt e)~\sf\dfrac{x}{5}=21-\dfrac{x}{2}\cdot(10)\\\sf 2x=210-5x\\\sf 2x+5x=210\\\sf 7x=210\\\sf x=\dfrac{210}{7}\\\\\sf x=30\\\tt f)~\sf\dfrac{4}{5}+\dfrac{3x}{4}=\dfrac{1}{10}+x\cdot(20)\\\\\sf16+15x=2+20x\\\sf 15x-20x=2-16\\\sf -5x=-14\cdot(-1)\\\sf 5x=14\\\\\sf x=\dfrac{14}{5}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Grupo~4}\\\tt a)~\sf\dfrac{2\cdot(4x-3)}{3}-\dfrac{4\cdot(4x+5)}{9}-\dfrac{5\cdot(x-4)}{6}=2\cdot(18)\\\\\sf 12\cdot(4x-3)-8\cdot(4x+5)-15\cdot(x-4)=36\\\sf 48x-36-32x-40-15x+60=36\\\sf 48x-32x-15x=36+36+40-60\\\sf x=52\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf\dfrac{2x+5}{3}-\dfrac{4x-9}{6}=\dfrac{3-4x}{2}\cdot(6)\\\sf2\cdot(2x+5)-1\cdot(4x-9)=3\cdot(3-4x)\\\sf \diagdown\!\!\!\!\!4x+10-\diagdown\!\!\!\!\!\!4x+\backslash\!\!\!9=\backslash\!\!\!9-12x\\\sf 12x=-10\\\sf x=-\dfrac{10\div2}{12\div2}\\\\\sf x=-\dfrac{5}{6}\end{array}}$