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Sunny_sXe [5.5K]
3 years ago
9

How do i solve the linear system -14x+15y=15 21x-20y=-10 using elimination

Mathematics
1 answer:
valina [46]3 years ago
4 0
Solve the following system using elimination:
{15 y - 14 x = 15 | (equation 1)
{21 x - 20 y = -10 | (equation 2)

Swap equation 1 with equation 2:
{21 x - 20 y = -10 | (equation 1)
{-(14 x) + 15 y = 15 | (equation 2)

Add 2/3 × (equation 1) to equation 2:
{21 x - 20 y = -10 | (equation 1)
{0 x+(5 y)/3 = 25/3 | (equation 2)

Multiply equation 2 by 3/5:
{21 x - 20 y = -10 | (equation 1)
{0 x+y = 5 | (equation 2)

Add 20 × (equation 2) to equation 1:
{21 x+0 y = 90 | (equation 1)
{0 x+y = 5 | (equation 2)

Divide equation 1 by 21:
{x+0 y = 30/7 | (equation 1)
{0 x+y = 5 | (equation 2)

Collect results:

Answer:  {x = 30/7, y = 5
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1. -88 = -3(4m + 5) - (1 - 3m)​
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Answer:

m=8

Step-by-step explanation:

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Express each number in scientific notation. <br><br> 18<br><br><br> 8,888,800
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What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form? Quadratic formula: x = StartFraction neg
Harman [31]

Answer:

x=\frac{-5(+/-)\sqrt{5}} {2}

x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

f(x)=x^{2} +5x+5  

equate the function to zero

x^{2} +5x+5=0  

so

a=1\\b=5\\c=5

substitute in the formula

x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(5)}} {2(1)}

x=\frac{-5(+/-)\sqrt{5}} {2}

therefore

x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction

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3 years ago
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