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Solve this system of equations
X=y+2
2y-3x=15

Sagot :

Answer:

x=-19, y=-21

Step-by-step explanation:

[tex]\begin{bmatrix}x=y+2\\ 2y-3x=15\end{bmatrix}[/tex]

[tex]\mathrm{Substitute\:}x=y+2[/tex]

[tex]\begin{bmatrix}2y-3\left(y+2\right)=15\end{bmatrix}[/tex]

Simplify

[tex]\begin{bmatrix}-y-6=15\end{bmatrix}[/tex]

Isolate y for -y - 6 = 15: y = -21

[tex]\mathrm{For\:}x=y+2[/tex]

[tex]\mathrm{Substitute\:}y=-21[/tex]

[tex]x=-21+2[/tex]

Simplify

x=-19

[tex]\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}[/tex]

[tex]x=-19,\:y=-21[/tex]

Jamech

y=-21,x=-19

you can only solve this using simultaneouse solution. first, make x the subject in the second equation.2y-3x=15

In doing so x=-5+2/3y. You the multiply both equationby the co-efficient of x then you subtract the equations giving 1/3y=-7. hence y=-21.you then substitute the value for y into equation 1 where x=y+2 giving x=-19.

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