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MArishka [77]
3 years ago
11

What is 6/7 - 2/5 please help

Mathematics
2 answers:
vazorg [7]3 years ago
7 0
LCD=35.so 30/35-14/35=16/35.welcome

guajiro [1.7K]3 years ago
4 0
6/7-2/5 -> Find LC denominator ->35

7x5=35 so multiply num and den of 6/7 by 5

5x7=35 so multiply num and den of 2/5 buy 7

Works out to: 30/35 - 14/35 

30-14= 16  answer is 16/35
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I need answer on this graphing va substitution.Problem 7 I tried to solve it but not sure that’s correct
Bond [772]

Given the system of equations:

\begin{gathered} 2x+9y=27 \\ x-3y=-24 \end{gathered}

To solve it by substitution, follow the steps below.

Step 1: Solve one linear equation for x in terms of y.

Let's choose the second equation. To solve it for x, add 3y to each side of the equations.

\begin{gathered} x-3y=-24 \\ x-3y+3y=-24+3y \\ x=-24+3y \end{gathered}

Step 2: Substitute the expression found for x in the first equation.

\begin{gathered} 2x+9y=27 \\ 2\cdot(-24+3y)+9y=27 \\ -48+6y+9y=27 \\ -48+15y=27 \end{gathered}

Step 3: Isolate y in the equation found in step 2.

To do it, first, add 48 to both sides.

\begin{gathered} -48+15y=27 \\ -48+15y+48=27+48 \\ 15y=75 \end{gathered}

Then, divide both sides by 15.

\begin{gathered} \frac{15y}{15}=\frac{75}{15} \\ y=5 \end{gathered}

Step 4: Substitute y by 5 in the relation found in step 1 to find x.

\begin{gathered} x=-24+3y \\ x=-24+3\cdot5 \\ x=-24+15 \\ x=-9 \end{gathered}

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x = -9

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or (-9, 5)

Also, you can graph the lines by choosing two points from each equation, according to the picture below.

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1 year ago
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\bf \left. \qquad  \right.\textit{internal division of a line segment}
\\\\\\
A(5,0)\qquad C(x,y)\qquad
\qquad 2:5
\\\\\\
\cfrac{AB}{BC} = \cfrac{2}{5}\implies \cfrac{A}{C}=\cfrac{2}{5}\implies 5A=2C\implies 5(5,0)=2(x,y)\\\\
-------------------------------\\\\
{ B=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}

\bf -------------------------------\\\\
B=\left(\cfrac{(5\cdot 5)+(2\cdot x)}{2+5}\quad ,\quad \cfrac{(5\cdot 0)+(2\cdot y)}{2+5}\right)=\boxed{(8,0)}
\\\\\\
B=\left( \cfrac{25+2x}{7}~~,~~\cfrac{0+2y}{7} \right)=(8,0)\implies 
\begin{cases}
\cfrac{25+2x}{7}=8\\\\
25+2x=56\\
2x=31\\\\
\boxed{x=\cfrac{31}{2}}\\
-------\\
\cfrac{0+2y}{7}=0\\\\
2y=0\\
\boxed{y=0}
\end{cases}
8 0
3 years ago
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