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neonofarm [45]
2 years ago
5

X+y=7 x-y=5 solution with subsitution method

Mathematics
2 answers:
pentagon [3]2 years ago
4 0

Topic : Linear equations in two variables

\:

Given Equations,

  • x + y = 7 ----- (i)

  • x - y = 5 ----- (ii)

\:

Solution,

First of all, let us take the first equation,

\\ \longrightarrow \qquad{  {{ \sf{ x + y}   = 7 \:  \:}}}  \\  \\ \:  \:

Subracting y from both sides we get :

\\  \longrightarrow \qquad{  {{ \sf{ x + y - y}   = 7 - y \:  \:}}}  \\  \:  \:

\longrightarrow \qquad{  {{ \sf{ x }   = 7 - y \:  \:}}}  \\  \\ \:  \:

Now, Substituting the value of x in Equation (ii) :

\\ \longrightarrow \qquad{  {{ \sf{ x-y=5 \:  \:}}}}  \\  \:  \:

\longrightarrow \qquad{  {{ \sf{ (7 - y)-y=5 \:  \:}}}}  \\  \:  \:

\longrightarrow \qquad{  {{ \sf{ 7  - 2y=5 \:  \:}}}}  \\  \\ \:  \:

Now, Subtracting 7 from both sides :

\\ \longrightarrow \qquad{  {{ \sf{ 7  - 2y - 7=5 - 7 \:  \:}}}}  \\ \:  \:

\longrightarrow \qquad{  {{ \sf{   - 2y = - 2\:  \:}}}}  \\  \\ \:  \:

Dividing both sides by -2 we get :

\\ \longrightarrow \qquad{  {{ \sf{  \frac{ - 2y}{ - 2} = \frac{ - 2}{ - 2} \:  \:}}}}  \\  \:  \:

\longrightarrow \qquad{ \underline{\boxed{ \pmb{ \mathfrak{y  =  \:  \: 1}}}}} \: \bigstar\\ \:

  • <u>Therefore, The value of y is 1 </u>

Now, Substituting the value of y in Equation (i) :

\\ \longrightarrow \qquad{  {{ \sf{ x+y=7 \:  \:}}}}  \\  \:  \:

\longrightarrow \qquad{  {{ \sf{ x+1=7 \:  \:}}}}  \\  \\ \:  \:

Subracting 1 from both sides we get :

\\ \longrightarrow \qquad{  {{ \sf{ x+1 - 1=7  - 1\:  \:}}}}  \\  \:  \:

\longrightarrow \qquad{ \underline {\boxed{ \pmb{ \mathfrak{x  =  \:  \: 6}}}}} \:  \: \bigstar\\  \:

  • <u>Therefore, The value of x is 16</u>
VladimirAG [237]2 years ago
3 0

Hey ! there

Answer:

Solution of equation or we can say that value of

  • x = <u>6</u>

  • y = <u>1</u>

Step-by-step explanation:

In this question we are given with <u>two equations that are ,</u>

  • <u>two equations that are ,x + y = 7</u>

  • <u>two equations that are ,x + y = 7x - y = 5</u>

And we are asked to <u>find the solution of equation with the help of substitution method .</u>

<u>SOLUTION</u><u> </u><u>:</u><u> </u><u>-</u>

Firstly we are giving numbering to the equation so that there's ease in solving . So ,

  • x + y = 7 ----------- <u>( Equation 1 ) </u>

  • x - y = 5 ----------- <u>( Equation 2 )</u>

We can see that Equation 2 is smaller than Equation 1 . So we are using Equation 2 to find the value of x .

<u>Finding value of x from Equation 2 :</u>

  • \rm{x - y = 5}

Adding y on both sides :

  • \rm{x -  \cancel{y }+ \cancel{ y }= 5  + y}

We get :

  • \underline{ \boxed{ \rm{x = 5 + y}}} -  -  -  - (\rm{Equation \: 3})

Therefore , value of x is <u>5</u><u> </u><u>+</u><u> </u><u>y </u><u>.</u>

Now substituting value of <u>x as 5 + y</u> in Equation 1 in order to find the <u>value of y.</u> So ,

\:  \quad \:  \longmapsto  \: \qquad \: \rm{ x + y = 7}

<u>S</u><u>t</u><u>e</u><u>p</u><u> </u>1 : Substituting value of x :

\:  \quad \:  \longmapsto  \: \qquad \: \rm{ \bold{5 + y} + y = 7}

<u>Step </u>2 : Adding like terms that are y and y :

\:  \quad \:  \longmapsto  \: \qquad \: \rm{5 + 2y = 7}

<u>Step </u>3 : Subtracting 5 on both sides :

\:  \quad \:  \longmapsto  \: \qquad \: \rm{ \cancel{5} + 2y -  \cancel{5} = 7 - 5}

We get ,

\:  \quad \:  \longmapsto  \: \qquad \: \rm{2y = 2}

<u>Step </u>4 : Dividing both sides with 2 :

\:  \quad \:  \longmapsto  \: \qquad \: \rm{ \dfrac{ \cancel{2}y}{ \cancel{2}}  =  \cancel{ \dfrac{2}{2} }}

On simplifying, We get :

\:  \quad \:  \longmapsto  \: \qquad \:  \blue{ \underline{\boxed{ \frak{y = 1}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>value </u><u>of </u><u>y </u><u>is </u><u>❝</u><u> </u><u>2</u><u> </u><u>❞</u>

<u>Now finding value of x </u><u>from </u><u>Equation</u><u> </u><u>3 </u><u>: </u>

<u>For </u><u>finding</u><u> </u><u>value </u><u>of </u><u>x </u><u>we </u><u>are </u><u>substituting</u><u> </u><u>value </u><u>of </u><u>y </u><u>in </u><u>Equation</u><u> </u><u>3 </u><u>.</u><u> </u><u>So </u><u>,</u>

<u>\:  \quad \:  \longmapsto  \: \qquad \: \rm{x = 5 + y}</u>

Substituting value of y :

\:  \quad \:  \longmapsto  \: \qquad \: \rm{x = 5 + 1}

Adding 5 with 1 , We get :

\:  \quad \:  \longmapsto  \: \qquad \:    \blue{\underline{\boxed{\frak{x = 6}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>value </u><u>of </u><u>x </u><u>is </u><u>❝</u><u> </u><u>6</u><u> </u><u>❞</u>

<u>From </u><u>values </u><u>of </u><u>x </u><u>as </u><u>6</u><u> </u><u>and </u><u>y </u><u>as </u><u>1</u><u> </u><u>we </u><u>can </u><u>say </u><u>that </u><u>they </u><u>are </u><u>the </u><u>solution</u><u> </u><u>of </u><u>given </u><u>equations</u><u> </u><u>.</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are checking our answer whether it is wrong or right .

<u>Equation</u><u> </u><u>1</u><u> </u><u>:</u><u> </u><u>x </u><u>+</u><u> </u><u>y </u><u>=</u><u> </u><u>7</u>

Substituting value<u> </u>of x and y in Equation 1 :

  • 6 + 1 = 7

  • 7 = 7

  • L.H.S = R.H.S

  • Hence, Verified.

<u>Therefore</u><u>,</u><u> our</u><u> answer</u><u> is</u><u> correct</u><u> </u><u>.</u>

<u>Equation 2 :</u><u> x</u><u> - y = 1</u>

  • 6 - 1 = 5

  • 5 = 5

  • L.H.S = R.H.S

  • Hence , Verified .

<u>Therefore</u><u> </u><u>,</u><u> </u><u>our</u><u> answer</u><u> is</u><u> correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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