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2 years ago

3x - 2y= 5, 4x + 3y= 1

Mathematics

2 answers:

Sindrei [870]2 years ago

7 0

Answer:

(1, - 1 )

Step-by-step explanation:

Given the 2 equations

3x - 2y = 5 → (1)

4x + 3y = 1 → (2)

Multiplying (1) by 3 and (2) by 2 and adding the result will eliminate y- term

9x - 6y = 15 → (3)

8x + 6y = 2 → (4)

Add (3) and (4) term by term to eliminate y

17x + 0 = 17

17x = 17 ( divide both sides by 17 )

x = 1

Substitute x = 1 into either of the 2 equations and solve for y

Substituting into (2)

4(1) + 3y = 1

4 + 3y = 1 ( subtract 4 from both sides )

3y = - 3 ( divide both sides by 3 )

y = - 1

solution is (1, - 1 )

Elena-2011 [213]2 years ago

6 0

Answer:

x=17 and y=23 DONT SUE ME IF I GET THIS WRONG PLZZ ;-;

Step-by-step explanation:

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$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

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$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

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3x - 2y= 5, 4x + 3y= 1​

3x - 2y= 5, 4x + 3y= 1