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Sever21 [200]
3 years ago
12

Can someone understand this i know its substitution for sure if i dont get wrong

Mathematics
1 answer:
svlad2 [7]3 years ago
7 0

4x - y = 3

y = 2x + 13


Substitute the second equation into the first equation:

4x - y = 3  [since y = 2x + 13, you can substitute (2x + 13) for "y" in the equation]

4x - (2x + 13) = 3    Distribute/multiply - into (2x + 13)

4x - 2x - 13 = 3  

2x - 13 = 3       Add 13 on both sides

2x = 16     Divide 2 on both sides

x = 8


Now that you know x = 8, plug it into one of the equations to find y

y = 2x + 13

y = 2(8) + 13

y = 16 + 13

y = 29


x = 8, y = 29  or  (8, 29)


[proof]

4x - y = 3

4(8) - 29 = 3

32 - 29 = 3

3 = 3


y = 2x + 13

29 = 2(8) + 13

29 = 16 + 13

29 = 29


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4 0
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Change the expression to a single square root, or its opposite:
aleksley [76]

Answer:

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e)-6\sqrt{2a}  =-\sqrt{72a}

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Step-by-step explanation:

a) 2\sqrt{2}=( \sqrt{2}  )^2.\sqrt{2}=(\sqrt{2})^3 = \sqrt{2^3} =\sqrt{8}

b)-7\sqrt{3} =-(\sqrt{7} )^2\sqrt{3} =-\sqrt{7^2.3} =-\sqrt{147}

c)\frac{1}{3} \sqrt{18b} =\frac{1}{3} \sqrt{9.2.b} =\frac{1}{3} \sqrt{3^2.2.b} =\frac{1}{3} \times 3\sqrt{2.b} =\sqrt{2.b}

d)5\sqrt{y} =\sqrt{5^2} \sqrt{y}=\sqrt{25y}

e)-6\sqrt{2a} =-\sqrt{6^2}\sqrt{2a}  = -\sqrt{36.2a} =-\sqrt{72a}

f)-0.1\sqrt{200c}=-\frac{1}{10}  \sqrt{10^2.2c} =-\frac{1}{10}\times10  \sqrt{2c}=-  \sqrt{2c

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