<p>i have been trying to solve this for a while today, for almost 1hr.
it's not my homework, but i saw the question with the triangle posted in SAT I math questions</p>
<p>a+x=130
x+y=140
b+y=160
b+a=150</p>
<p>please solve for all the variables, please help</p>
<p>use a system of equations</p>
<p>a = 60, b = 90, x = 70, y = 70</p>
<p>I just thought about it and plugged in the right numbers.</p>
<p>Is the answer x=20, a =110, b= 40, y=120?</p>
<p>I just plugged in numbers less than 130.</p>
<p>null set? :o</p>
<p>There is definitely something missing. If given a value for x+b or y+a, it is possible, but now x+20=b and a+10=y. As long as those are true, the value of each can be anything.</p>
<p>a=60
b=90
x=70
y=70</p>
<p>I tried both of these and they both worked:</p>
<p>a=80
y=90
x=50
b=70</p>
<p>a=70
y=80
x=60
b=80</p>
<p>Thus proving my point.</p>
<p>I'll show you how to solve it with matricies.</p>
<p>Set up Matrix 1
I will label the columns for each variable, and when that variable is present put in the coefficent (e.g. a=1, 2b=2, 3b^2=9). and multiply that matrix by the variables in order (a,b,x,y</p>
<p>[A B X Y]
[1 0 1 0] [a]
[0 0 1 1] **
[0 1 0 1] [c]
[1 1 0 0] [d]</p>
<p>Find the Inverse matrix of Matrix 1. The inverse matrix when multiplied by the origonal will result in the a matrix that is communitve. Then setup your second matrix in the order that you set up the first matrix, if you dont you will get a wrong answer!!</p>
<p>[130]
[140]
[160]
[150]</p>
<p>Now the equation looks like
[matrix 1][a] = [130]
** = [140]
[x]= [160]
[y]= [150]</p>
<p>Take that inverse matrix we just solved for and multiply that matrix by this set of matrices and you have the answers in order for a,b,x,y.</p>
<p>[matrix 1] * [inverse] = communitive matrix, which leaves a,b,x,and y the
same</p>
<p>The answer will be
[a]= [inverse of matrix 1] * [130]
**= [140]
[x]= [160]
[y]= [150]</p>
<p>There are infinitely many solutions to the system. The solutions are in the form:</p>
<p>a=150-s
b=s
x=s-20
y=160-s
where s is any real number</p>
<p>In reference to ryfrky07, that method does not work here because the coefficient matrix is singular.</p>
<p>I agree with MetsFan. We can't solve it without additional information. Was there a diagram of the triangle? Was one of the variables an exterior angle? Something's definitely missing. </p>
<p>Also, I found another solution set. </p>
<p>a = 50
x = 80
y = 60
b = 100</p>
<p>If you have n independent linear equations in n variables, you can solve them by matrix algebra. However, the 4 equations shown here are NOT independent; Eqn(1) - Eqn(2) + Eqn(3) yields the same info shown in Eqn(4). </p>
<p>So (as AverageMathGeek said in post#10) you have an infinite #solutions here; discard equation(4), assume any value you want for (say) a, then solve the remaining (3x3) system of equations for values of b,x,y.</p>
<p>If the determinant of the matrix of the coefficients is zero, the system is either dependent (infinite number of solutions) or inconsistent (no solutions). </p>
<p>det(
[1 0 1 0]
[0 0 1 1]
[0 1 0 1]
[1 1 0 0] ) = 0.</p>
<p>so it's not just me who can't solve this? b/c this is a impossible question right?</p>
<p>You are right if the question was presented in exactly this form.
+++++++++++++++
Interestingly enough, a system can be dependent with the ifinite number of sets of numbers as solutions, but some variables stay costant.
For example, this system:
x-y=2
2x-2y=4
x-y+z=0.
Solutions: (k, k-2, 2), where k - any number.</p>
<p>I meant (k, k-2, -2)</p>