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alekssr [168]
2 years ago
10

Solve the system by substitution. 4x + 2y = 10 x − y = 13

Mathematics
2 answers:
Rainbow [258]2 years ago
8 0

answer are attached in pic hope it's helpful for you

value of x =6

y =7

ziro4ka [17]2 years ago
5 0

Answer:

\boxed{\tt (6,-7)}

Step-by-step explanation:

\tt 4x + 2y = 10\\x - y = 13

First Let's solve for x in x-y=13

\tt x-y=13

Add y to both sides:

\tt x=13+y

Now, Substitute x=13+y into 4x+2y=10:-

Let x=13+y

\tt 4(13+y)+2y=10

\tt 52+6y=10

Now, let's solve for y in 52+6y=10:-

\tt 52+6y=10

Subtract 52 from both sides:-

\tt 6y=10-52

\tt 6y=-42

Divide both sides by 6:-

\tt \cfrac{6y}{6} =\cfrac{-42}{6}

\boxed{\tt y=-7}

Now, we'll Substitute y=-7 into to x=13+y to find "x":-

\tt x=13+y

\tt x=13+-7

\boxed{\tt x=6}

___

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Which of the following are solutions to the equation below?
scoundrel [369]

Answer:

The correct options for the solution values are:

  • x=2\sqrt{2}-5
  • x=-2\sqrt{2}-5

Step-by-step explanation:

Given the expression

x^2+10x+25=8

Subtract 25 from both sides

x^2+10x+25-25=8-25

Simplify

x^2+10x=-17

Add 25 or 5² to both sides

x^2+10x+5^2=8

as

x^2+10x+5^2=\left(x+5\right)^2

so the expression becomes

\left(x+5\right)^2=8

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solve

x+5=\sqrt{8}

Subtract 5 from both sides

x+5-5=2\sqrt{2}-5

x=2\sqrt{2}-5

solve

x+5=-\sqrt{8}

Subtract 5 from both sides

x+5-5=-2\sqrt{2}-5

x=-2\sqrt{2}-5

Therefore, the solution to the equation

x=2\sqrt{2}-5,\:x=-2\sqrt{2}-5

Hence, the correct options for the solution values are:

  • x=2\sqrt{2}-5
  • x=-2\sqrt{2}-5

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Step-by-step explanation:

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