Hobidid321
Answered

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Solve the following simulataneous equations using an algebraic method (not graphical).
2x + 5y = 0
3x-4y = 23

Sagot :

Given:

The equations are

[tex]2x+5y=0[/tex]

[tex]3x-4y=23[/tex]

To find:

The solution of given system of equations.

Solution:

We have,

[tex]2x+5y=0[/tex]        ...(i)

[tex]3x-4y=23[/tex]       ...(ii)

From (i), we get

[tex]2x=-5y[/tex]

[tex]x=-\dfrac{5y}{2}[/tex]        ...(iii)

Putting this value in (ii), we get

[tex]3(-\dfrac{5y}{2})-4y=23[/tex]

[tex]-\dfrac{15y}{2}-4y=23[/tex]

[tex]\dfrac{-15y-8y}{2}=23[/tex]

[tex]-23y=2\times 23[/tex]

[tex]y=\dfrac{46}{-23}[/tex]

[tex]y=-2[/tex]

Putting y=-2 in (iii), we get

[tex]x=-\dfrac{5(-2)}{2}[/tex]

[tex]x=-\dfrac{-10}{2}[/tex]

[tex]x=\dfrac{10}{2}[/tex]

[tex]x=5[/tex]

Therefore, the solution of given system of equation is (5,-2).

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