Solve by using substitution
x=3y-1
x+2y=9

x=3y−1x=3y-1

x+2y=9x+2y=9

Replace all occurrences of xx with 3y−13y-1 in each equation.

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Replace all occurrences of xx in x+2y=9x+2y=9 with 3y−13y-1.

(3y−1)+2y=9(3y-1)+2y=9

x=3y−1x=3y-1

Add 3y3y and 2y2y.

5y−1=95y-1=9

x=3y−1x=3y-1

Solve for yy in the first equation.

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Move all terms not containing yy to the right side of the equation.

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Add 11 to both sides of the equation.

5y=9+15y=9+1

x=3y−1x=3y-1

Add 99 and 11.

5y=105y=10

x=3y−1x=3y-1

Divide each term by 55 and simplify.

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Divide each term in 5y=105y=10 by 55.

5y5=1055y5=105

x=3y−1x=3y-1

Cancel the common factor of 55.

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Cancel the common factor.

5y5=1055y5=105

x=3y−1x=3y-1

Divide yy by 11.

x=5,y=2

Step-by-step explanation:

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option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point