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3 years ago

3x + 4y = 27 5x - 3y = 16

Mathematics

2 answers:

maks197457 [2]3 years ago

5 0

Answer:

x = 5 , y = 3

Step-by-step explanation:

Solve the following system:

{3 x + 4 y = 27 | (equation 1)

5 x - 3 y = 16 | (equation 2)

Swap equation 1 with equation 2:

{5 x - 3 y = 16 | (equation 1)

3 x + 4 y = 27 | (equation 2)

Subtract 3/5 × (equation 1) from equation 2:

{5 x - 3 y = 16 | (equation 1)

0 x+(29 y)/5 = 87/5 | (equation 2)

Multiply equation 2 by 5/29:

{5 x - 3 y = 16 | (equation 1)

0 x+y = 3 | (equation 2)

Add 3 × (equation 2) to equation 1:

{5 x+0 y = 25 | (equation 1)

0 x+y = 3 | (equation 2)

Divide equation 1 by 5:

{x+0 y = 5 | (equation 1)

0 x+y = 3 | (equation 2)

Collect results:

Answer: {x = 5 , y = 3

Serhud [2]3 years ago

3 0

Answer:

The lines will cross at (5,3)

Step-by-step explanation:

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3 years ago

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Which expression is equivalent to line l-5l + l3l? -8 -2 2 8

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1.) through: (3,0), parallel to y=2/3x+1

ruslelena [56]

Step-by-step explanation:

Hey there!

1.no Ans:

The equation of a st.line passing through point (3,0) is;

(y-0) = m1(x-3)........(i)

Another equation is,

y = 2/3x+1...........(ii)

Comparing the equation (ii) with y= mx+c. We get;

Slope (m2)= 2/3

As they are parallel lines, m1=m2.

So, putting value of m1 in equation (i).

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

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

Therefore the required equation is; y = 2/3x -2.

2.Ans:

The equation of a st.line passing through (4,1) is;

(y-1) = m1(x-4)...........(i)

y = -1/2x +2...............(ii)

From equation (ii)

y = -1/2x+2

Comparing the equation with y = mx+c

Slope (m2)= -1/2

As they are equal m1=m2= -1/2

Putting value of slope in equation (i)

$(y - 1) = \frac{ - 1}{2} (x + 2)$

$2(y - 1) = - 1(x + 2)$

$2y - 2 = - x - 2$

$x + 2y = 0$

Therefore the required equation is; x+2y=0.

3. no. Ans:

The equation of a st.line passing through point (-1,-1) is ;

(y+1)= m1(x+1)..........(i)

y= -2x-4.....................(ii)

From equation (ii)

Comparing the equation with y = mx+c.

Slope(m2)= -2

As they are parallel lines, m1=m2 = -2.

Putting value of slope in equation (i)

$(y + 1) = - 2(x + 1)$

$y + 1 = - 2x - 2$

$2x + y + 3 = 0$

Therefore the required equation is 2x+y+3=0.

4.no Ans:

The equation of a st.line passing through (-4,-5) is;

(y+5) = m1(x+4)..........(i)

y = -2x-5...............(ii)

From equation (ii)

Comparing the equation with y = mx+c, we get;

Slope (m2)= -2

As they are parallel lines, m1=m2= -2

Putting value of slope in equation (i)

$(y + 5) = - 2(x + 4)$

$y + 5 = - 2x - 8$

$2x + y + 13 = 0$

Therefore the required equation is 2x+y+13=0.

5.no Ans:

The equation of a st.line passing through (-4,3) is:

(y-3)= m1(x+4).........(i)

y = 1/2x-3.......(ii)

From equation (ii)

Comparing the equation with y= mx+c.

Slope (m2) = 1/2

As they are parallel lines, m1=m2 = 1/2.

Putting value of slope in equation (i)

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

$2(y - 3) = x + 4$

$2y - 6 = x + 4$

$x - 2y + 10 = 0$

Therefore the required equation is; x-2y+10=0.

6.no. Ans;

The equation of a st.line passing through (5,-5) is;

(y+5)= m1(x-5)............(i)

y = -3/2x+2......(ii)

Comparing the equation (ii) with y = mx+c.

Slope(m2)= -3/2.

As they are parallel lines, m1=m2= -3/2.

Putting value of slope in equation (i)

$(y + 5) = \frac{ - 3}{2} (x - 5)$

$2(y + 5) = - 3(x - 5)$

$2y + 10 = - 3x + 15$

$3x + 2y - 5 = 0$

Therefore the required equation is; 3x+2y-5=0.

Hope it helps...

7 0

3 years ago