Solve the system by the elimination method. Check your work. 3a + 5b - 7 = 0 a - 2b - 4 = 0

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Solve the system by the elimination method. Check your work. 3a + 5b - 7 = 0 a - 2b - 4 = 0

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find the LCM for one of your variables and then subtract the equations from eachother
he first one is Let's solve for a. 3a+5b−7=0 Step 1: Add 7 to both sides. 3a+5b−7+7=0+7 3a+5b=7 Step 2: Add -5b to both sides. 3a+5b+−5b=7+−5b 3a=−5b+7 Step 3: Divide both sides by 3. 3a 3 = −5b+7 3 a= −5 3 b+ 7 3 Answer: a= −5 3 b+ 7 3
Let's solve for a. 3a+5b−7=0 Step 1: Add 7 to both sides. 3a+5b−7+7=0+7 3a+5b=7 Step 2: Add -5b to both sides. 3a+5b+−5b=7+−5b 3a=−5b+7 Step 3: Divide both sides by 3. 3a 3 = −5b+7 3 a= −5 3 b+ 7 3 Answer: a= −5 3 b+ 7 3

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A. {(96/11, -5/11)} B. {(34/11, -5/11)} C. {(32/33, 5/11)} These are my choices.
well 1(3a+5b-7=0) -3(a-2-4=0 3a+5b-7=0 -3a+6b+12=-3 11+4b=-3 then 4b=8 4/8 2=b So after that you will input into equation a - 2(2) - 4 = 0 a-4-4=o a-0=0 a=0 is what i belive it would be so, 2=b 0=a
{(34/11, -5/11)}
@Brandon1001 Did you check to see if those answers work? I did not get that but I have to check my work.
is the answer
Solve the system by the elimination method. Check your work. 3a + 5b - 7 = 0 a - 2b - 4 = 0 I would multiply the bottom one by -3 -3a+6b+12=0 3a+5b-7=0 <----top equation The a's cancel out and we are left with 11b+5=0 11b=-5 \[b=-\frac{ 5 }{ 11 }\] Now plug that in to find a \[a-2(-\frac{ 5 }{ 11 })-4=0\] \[a+\frac{ 10 }{ 11 }-4=0\] \[a+\frac{ 10 }{ 11 }-\frac{ 44 }{ 11 } =0\] \[a-\frac{ 34 }{ 11 }=0\] \[a=\frac{ 34 }{ 11 }\]
So {(34/11, -5/11)} is correct

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