Question

24. Solve for x and y, if 2x + 3y = 8 and x - 2y + 3 = 0​

Answer 1

Step-by-step explanation:

GivenLinearpairofequations:

2x+3y = 8 \: --(1)2x+3y=8−−(1)

\begin{gathered}\begin {tabular}{|c|c|c|}\cline{1-3} x&-2&4\\\cline{1-3} y&4&0\\\cline{1-3} (x,y)&A(-2,4)&B(4,0)\\\cline{1-3} \end {tabular}\end{gathered}

and \: x - 2y + 3 = 0 \: --(2)andx−2y+3=0−−(2)

\begin{gathered}\begin {tabular}{|c|c|c|}\cline{1-3} x&-1&3\\\cline{1-3} y&-2&0\\\cline{1-3} (x,y)&C(-1,-2)&D(3,0)\\\cline{1-3} \end {tabular}\end{gathered}

Now , Plotting the points A,B on the graph and joining them we get a straight line .

And , Plotting the points C,D on the graph and joining them we get another straight line .

These two straight lines intersect at point P.

Finding Coordinates of point P :

2x + 3y = 8 \: --(1)2x+3y=8−−(1)

and \: x = 3 + 2y \: ---(2)andx=3+2y−−−(2)

/* Put x = 3 + 2y in equation (1) , we get */

2(3+2y) + 3y = 82(3+2y)+3y=8

\implies 6 + 4y + 3y = 8⟹6+4y+3y=8

\implies 7y = 8 - 6⟹7y=8−6

\implies y = \frac{2}{7}⟹y=

7

2

Now , value \: of \: y \: in /:equation \: (2) , we \:getNow,valueofyin/:equation(2),weget

x = 3 + 2 \times \frac{2}{7}x=3+2×

7

2

\implies x = \frac{21+4}{7}⟹x=

7

21+4

\implies x = \frac{25}{7}⟹x=

7

25

\red{Required \: Solution }RequiredSolution

\green { = P\Big(\frac{25}{7} , \frac{2}{7}\Big)

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