HELP ME PLEASE!!

Solve the simultaneous equations.
x + 2y = 13
x + 5y = 28

x =
y= ​

Answers

Answer 1

[tex]\bold{\huge{\underline{ Solution }}}[/tex]

Given :-

Here, We have given two equations that is ,[tex]\sf{ x + 2y = 13\:\: and\:\: x + 5y = 28}[/tex]

To Find :-

We have to find the value of x and y.

Concept used :-

Linear equations are those equations which having highest power of degree as 1 .Linear equations can be solved by subsitute method, elimination method and cross multiplication method. Here, I have used substitution method to make calculation easier. In subsitute method, you have to subsitute the value of one variable of eq(1) in eq(2) .

Let's Begin :-

Here,

We have two equations, let consider these two equations as eq(1) and eq(2) that is,

[tex]\sf{ x + 2y = 13 ...eq(1)}[/tex]

[tex]\sf{ x + 5y = 28 ...eq(2)}[/tex]

By solving eq(1) :-

[tex]\sf{ x + 2y = 13 }[/tex]

[tex]\sf{ x = 13 - 2y ...eq(3)}[/tex]

Subsitute eq(3) in eq(2) :-

[tex]\sf{ (13 - 2y) + 5y = 28 }[/tex]

[tex]\sf{ 13 - 2y + 5y = 28 }[/tex]

[tex]\sf{ 13 + 3y = 28 }[/tex]

[tex]\sf{ 3y = 28 - 13 }[/tex]

[tex]\sf{ 3y = 28 - 13 }[/tex]

[tex]\sf{ 3y = 15 }[/tex]

[tex]\sf{ y = }{\sf{\dfrac{ 15}{3}}}[/tex]

[tex]\sf{ y = }{\sf{\cancel{\dfrac{ 15}{3}}}}[/tex]

[tex]\bold{ y = 5 }[/tex]

Thus, The value of y is 5

Now,

Subsitute the value of y in eq(1) :-

[tex]\sf{ x + 2(5) = 13 }[/tex]

[tex]\sf{ x + 10 = 13 }[/tex]

[tex]\sf{ x = 13 - 10 }[/tex]

[tex]\bold{ x = 3 }[/tex]

Hence, The value of x and y are 5 and 3 .

[tex]\bold{\huge{\underline{ Let's \: Verify }}}[/tex]

Equation 1 :-

[tex]\sf{ x + 2y = 13 }[/tex]

[tex]\sf{ 3 + 2(5) = 13 }[/tex]

[tex]\sf{ 3 + 10 = 13 }[/tex]

[tex]\sf{ 13 = 13 }[/tex]

[tex]\bold{ LHS = RHS }[/tex]

Equation 2 :-

[tex]\sf{ x + 5y = 28 }[/tex]

[tex]\sf{ 3 + 5(5) = 28 }[/tex]

[tex]\sf{ 3 + 25 = 28 }[/tex]

[tex]\sf{ 28 = 28 }[/tex]

[tex]\bold{ LHS = RHS }[/tex]

Answer 2
x+2y=13--(1)x+5y=28--(2)

Let's use elimination method

Subtract eq(2) from eq(1)

We get

[tex]\\ \rm\Rrightarrow 2y-5y=13-28[/tex]

[tex]\\ \rm\Rrightarrow -3y=-15[/tex]

[tex]\\ \rm\Rrightarrow y=\dfrac{-15}{-3}[/tex]

[tex]\\ \rm\Rrightarrow y=5[/tex]

Putting on eq(1)

[tex]\\ \rm\Rrightarrow x+2y=13[/tex]

[tex]\\ \rm\Rrightarrow x+2(5)=13[/tex]

[tex]\\ \rm\Rrightarrow x+10=13[/tex]

[tex]\\ \rm\Rrightarrow x=13-10[/tex]

[tex]\\ \rm\Rrightarrow x=3[/tex]

(x,y)=(3,5)

[tex]\LARGE{\underbrace{\underline{\rm{Verification:-}}}}[/tex]

[tex]\\ \rm\Rrightarrow x+2y=13[/tex]

[tex]\\ \rm\Rrightarrow 3+2(5)=13[/tex]

[tex]\\ \rm\Rrightarrow 3+10=13[/tex]

[tex]\\ \rm\Rrightarrow 13=13[/tex]

And

[tex]\\ \rm\Rrightarrow x+5y=28[/tex]

[tex]\\ \rm\Rrightarrow 3+5(5)=28[/tex]

[tex]\\ \rm\Rrightarrow 3+25=28[/tex]

[tex]\\ \rm\Rrightarrow 28=28[/tex]

Hence verified!


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Step-by-step explanation:

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Step-by-step explanation:

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[tex]\tt -3x=6[/tex]

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[tex]\tt x=-2\;\; \bf Answer[/tex]

____

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[tex] \sf:\implies \quad {\pmb{\sf x = -2}} [/tex]

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Let's solve the given problem step by step :

[tex]:\implies \quad \sf - 3(x + 5) = - 9[/tex]

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[tex]:\implies \quad \sf - 3(x + 5) = - 9[/tex]

[tex] \sf:\implies \quad - (3 \times x) - (3 \times 5) = - 9[/tex]

Add 15 to both sides :

[tex] \sf:\implies \quad - 3x - 15 = - 9[/tex]

[tex] \sf:\implies \quad - 3x - 15 + 15 = - 9 + 15[/tex]

On simplifying :

[tex] \sf:\implies \quad -3x-15+15 = -9+15[/tex]

[tex] \sf:\implies \quad - 3x = 6[/tex]

Dividing both sides with same factor :

[tex] \sf:\implies \quad - 3x = 6[/tex]

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