Anyone help me pls,find the area of each figure(All lines meet at right angles.)
1 answer:
You might be interested in
3x-2y=6
1) <span>Add -3x to both sides.
</span><span><span> 3x</span>−<span>2y</span></span>=<span>6
</span>+3x +3x
-----------------
<span>−<span>2y</span></span>=<span><span>−<span>3x</span></span>+<span>6
</span></span><span>
2) Divide both sides by -2.
</span>
= 
y=<span><span><span>32</span>x</span>−<span>3
</span></span>
3) Substitute the y with the equation
3x-2(32x-3)=6
3x-64x+6=6
4) Combine like terms
3x-64x+6=6
-61x+6=6
5) Subtract 6 from each side of the equation
-61x+6=6
-6 -6 (+6-5 gets crossed out)
--------------
-61x=0
6) Divide both sides by -61
x=0
============================================================
Now that you've got you x intercept, use the x value and substitute it into the
y=32x−3
y=32(0)-3
y=0-3
y=-3
(0,-3)
1) <span>Add -3x to both sides.
</span><span><span> 3x</span>−<span>2y</span></span>=<span>6
</span>+3x +3x
-----------------
<span>−<span>2y</span></span>=<span><span>−<span>3x</span></span>+<span>6
</span></span><span>
2) Divide both sides by -2.
</span>
y=<span><span><span>32</span>x</span>−<span>3
</span></span>
3) Substitute the y with the equation
3x-2(32x-3)=6
3x-64x+6=6
4) Combine like terms
3x-64x+6=6
-61x+6=6
5) Subtract 6 from each side of the equation
-61x+6=6
-6 -6 (+6-5 gets crossed out)
--------------
-61x=0
6) Divide both sides by -61
x=0
============================================================
Now that you've got you x intercept, use the x value and substitute it into the
y=32x−3
y=32(0)-3
y=0-3
y=-3
(0,-3)