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2 years ago

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Solve:........................

x%20-%201%7D%7B5%7D%20" id="TexFormula1" title=" \frac{x}{3} + 4 = \frac{4x - 1}{5} " alt=" \frac{x}{3} + 4 = \frac{4x - 1}{5} " align="absmiddle" class="latex-formula">

1 answer:

lakkis [162]2 years ago

7 0

Answer:

$\frac{x}{3} + 4 = \frac{4x - 1}{5} \\ \\ \frac{x}{3} + \frac{12}{3} = \frac{4x - 1}{5} \\ \\ \frac{x + 12}{3} = \frac{4x - 1}{5} \\ \\ 3(4x - 1) = 5(x + 12) \\ \\ 12x - 3 = 5x + 60 \\ \\ 12x - 5x = 60 + 3 \\ \\ 7x = 63 \\ \\ x = \frac{63}{7} \\ \\ x = 9$

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Answer:

Part 1) $9x-7y=-25$

Part 2) $2x-y=2$

Part 3) $x+8y=22$

Part 4) $x+8y=35$

Part 5) $3x-4y=2$

Part 6) $10x+6y=39$

Part 7) $x-5y=-6$

Part 8)

case A) The equation of the diagonal AC is $x+y=0$

case B) The equation of the diagonal BD is $x-y=0$

Step-by-step explanation:

Part 1)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M=(\frac{2-6}{2},\frac{-3+5}{2})$

$M=(-2,1)$

step 2

The equation of the line into point slope form is equal to

$y-1=\frac{9}{7}(x+2)\\ \\y=\frac{9}{7}x+\frac{18}{7}+1\\ \\y=\frac{9}{7}x+\frac{25}{7}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=\frac{9}{7}x+\frac{25}{7}$

Multiply by 7 both sides

$7y=9x+25$

$9x-7y=-25$

Part 2)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M=(\frac{1+5}{2},\frac{0-2}{2})$

$M=(3,-1)$

step 2

Find the slope

The slope between two points is equal to

$m=\frac{-2-0}{5-1}=-\frac{1}{2}$

step 3

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

Find the slope of the line perpendicular to the segment joining the given points

$m1=-\frac{1}{2}$

$m1*m2=-1$

therefore

$m2=2$

step 4

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=2$ and point $(1,0)$

$y-0=2(x-1)\\ \\y=2x-2$

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=2x-2$

$2x-y=2$

Part 3)

In this problem AB and BC are the legs of the right triangle (plot the figure)

step 1

Find the midpoint AB

$M1=(\frac{-5+1}{2},\frac{5+1}{2})$

$M1=(-2,3)$

step 2

Find the midpoint BC

$M2=(\frac{1+3}{2},\frac{1+4}{2})$

$M2=(2,2.5)$

step 3

Find the slope M1M2

The slope between two points is equal to

$m=\frac{2.5-3}{2+2}=-\frac{1}{8}$

step 4

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{8}$ and point $(-2,3)$

$y-3=-\frac{1}{8}(x+2)\\ \\y=-\frac{1}{8}x-\frac{1}{4}+3\\ \\y=-\frac{1}{8}x+\frac{11}{4}$

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=-\frac{1}{8}x+\frac{11}{4}$

Multiply by 8 both sides

$8y=-x+22$

$x+8y=22$

Part 4)

In this problem the hypotenuse is AC (plot the figure)

step 1

Find the slope AC

The slope between two points is equal to

$m=\frac{4-5}{3+5}=-\frac{1}{8}$

step 2

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{8}$ and point $(3,4)$

$y-4=-\frac{1}{8}(x-3)$

$y=-\frac{1}{8}x+\frac{3}{8}+4$

$y=-\frac{1}{8}x+\frac{35}{8}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=-\frac{1}{8}x+\frac{35}{8}$

Multiply by 8 both sides

$8y=-x+35$

$x+8y=35$

Part 5)

The longer diagonal is the segment BD (plot the figure)

step 1

Find the slope BD

The slope between two points is equal to

$m=\frac{4+2}{6+2}=\frac{3}{4}$

step 2

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$ and point $(-2,-2)$

$y+2=\frac{3}{4}(x+2)$

$y=\frac{3}{4}x+\frac{6}{4}-2$

$y=\frac{3}{4}x-\frac{2}{4}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=\frac{3}{4}x-\frac{2}{4}$

Multiply by 4 both sides

$4y=3x-2$

$3x-4y=2$

Note The complete answers in the attached file

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3 years ago