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Solving for a Specific Variable Practice
Fifty original questions covering the complete solving for a specific variable lesson, with explanations and SAT-focused strategy.
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Question 1
Explanation
Treat \(A,B,C\) as constants. Subtract \(B\): \(Ax=C-B\). Divide the entire right side by \(A\): \(x=(C-B)/A\).
Question 2
Explanation
Subtract \(b\) from both sides: \(y-b=mx\). Divide both complete sides by \(m\): \(x=(y-b)/m\).
Question 3
Explanation
The target \(t\) is multiplied by \(r\). Divide both sides by \(r\): \(t=d/r\). Units also check: distance divided by distance per time equals time.
Question 4
Explanation
Width \(W\) is multiplied by \(L\). Dividing both sides by \(L\) gives \(W=A/L\).
Question 5
Explanation
Subtract \(2L\): \(P-2L=2W\). Divide the entire left side by \(2\): \(W=(P-2L)/2\), equivalently \(P/2-L\).
Question 6
Explanation
Multiply by \(9/5\): \(\frac95C=F-32\). Add \(32\): \(F=\frac95C+32\). The \(32\) is not inside the multiplication after the inverse step.
Question 7
Explanation
Subtract \(a\) from both sides: \(b=c-a\). The other letters remain symbolic constants.
Question 8
Explanation
Subtract \(p\): \(k-p=mn\). Divide by \(m\): \(n=(k-p)/m\). Both terms \(k-p\) remain in the numerator.
Question 9
Explanation
The phrase ‘for \(r\)’ explicitly names \(r\) as the target. Every other symbol is treated as a constant while \(r\) is isolated.
Question 10
Explanation
The target mindset treats every non-target symbol as a fixed quantity. This allows the same inverse-operation reasoning used in numerical equations.
Question 11
Explanation
The complete side \(C-B\) equals \(Ax\), so divide that entire side by \(A\): \(x=(C-B)/A\). Dividing only \(B\) is incomplete.
Question 12
Explanation
Without numerical values, rearranging to \(L=A/W\) is the complete answer. Literal-equation problems ask for a relationship, not an invented calculation.
Question 13
Explanation
A literal equation uses multiple symbols, often as a formula. Solving for one symbol rearranges the relationship while the others act as constants.
Question 14
Explanation
The complete group \(x-r\) is multiplied by \(k\). Divide both sides by \(k\) to get \(Q/k=x-r\), then add \(r\).
Question 15
Explanation
The final step divides both sides by \(A\), which is valid only when \(A\) is nonzero. If \(A=0\), the original equation must be classified separately.
Question 16
Explanation
Multiply by \(2\): \(2A=bh\). Divide by \(b\): \(h=2A/b\). Dividing by \(2b\) would halve instead of undoing the original one-half factor.
Question 17
Explanation
The complete coefficient of \(H\) is \(LW\). Divide both sides by that product: \(H=V/(LW)\).
Question 18
Explanation
The target \(r\) is multiplied by the product \(Pt\). Divide both sides by the complete product: \(r=I/(Pt)\).
Question 19
Explanation
Subtract \(u\): \(v-u=at\). Divide the complete difference by \(a\): \(t=(v-u)/a\).
Question 20
Explanation
Divide by \(k\): \(Q/k=x-r\). Add \(r\): \(x=Q/k+r\). The addition occurs after the outside factor has been undone.
Question 21
Explanation
Multiply by \(c\): \(Mc=a+b\). Subtract \(a\): \(b=Mc-a\). Clearing the denominator must happen before isolating the numerator term.
Question 22
Explanation
Multiply by \(s\): \(ps=q-r\). Add \(r\): \(q=ps+r\). The target sits inside the numerator, so clear the denominator first.
Question 23
Explanation
Multiply by \(t\): \(Rt=D\). Divide by \(R\): \(t=D/R\). Cross-multiplication here is just a compact form of those legal steps.
Question 24
Explanation
Multiply by \(x\): \(xy=a\). Divide by \(y\): \(x=a/y\). The original formula requires \(x\ne0\), and this rearrangement divides by \(y\).
Question 25
Explanation
Factor the target: \(P=x(r+s)\). Divide by the complete factor \(r+s\): \(x=P/(r+s)\).
Question 26
Explanation
Factor \(x\) on the right: \(M=x(3+y)\). Divide by \(3+y\): \(x=M/(3+y)\).
Question 27
Explanation
Collect: \(ax-cx=d-b\). Factor: \(x(a-c)=d-b\). Divide by \(a-c\): \(x=(d-b)/(a-c)\).
Question 28
Explanation
Subtract \(c\): \(K-c=x(a-b)\). Divide the entire left side by \(a-b\): \(x=(K-c)/(a-b)\).
Question 29
Explanation
Subtract \(rx\) and \(q\): \(px-rx=s-q\). Factor \(x(p-r)=s-q\), then divide by \(p-r\).
Question 30
Explanation
Subtract \(1/u\): \(1/v=1/f-1/u=(u-f)/(fu)\). Taking reciprocals gives \(v=fu/(u-f)\), with \(u\ne f\).
Question 31
Explanation
Subtract \(2W\): \(P-2W=2L\). Divide by \(2\): \(L=(P-2W)/2=P/2-W\).
Question 32
Explanation
Subtract \(32\): \(F-32=\frac95C\). Multiply by \(5/9\): \(C=\frac59(F-32)\). The difference remains grouped.
Question 33
Explanation
The complete coefficient of \(r\) is \(2\pi\). Divide both sides by \(2\pi\): \(r=C/(2\pi)\).
Question 34
Explanation
Multiply by \(2/h\): \(2A/h=b_1+b_2\). Subtract \(b_1\): \(b_2=2A/h-b_1\).
Question 35
Explanation
After subtracting \(B\), the equation is \(Ax=C-B\). Dividing by \(A\) applies to the entire right side, so \(x=(C-B)/A\).
Question 36
Explanation
Both right-side terms contain the target. Factoring gives \(P=x(r+s)\), so the full coefficient \(r+s\) must divide \(P\).
Question 37
Explanation
Multiplying both sides by \(x\) changes \(y=a/x\) to \(xy=a\), placing the target in an ordinary product that can be isolated by division.
Question 38
Explanation
Distribute division by \(2\) across the numerator: \((P-2L)/2=P/2-L\). Both forms correctly solve \(P=2L+2W\) for \(W\).
Question 39
Explanation
The formula divides by \(a-c\), so it requires \(a-c\ne0\). If \(a=c\), the equation instead becomes an identity or no-solution case depending on the constants.
Question 40
Explanation
Starting with \(Q=k(3x-r)+s\): subtract \(s\), divide by \(k\), add \(r\), and divide by \(3\). Combining the result gives \((Q-s+kr)/(3k)\).
Question 41
Explanation
Expand: \(A=px+pr-qx+qs\). Collect target terms: \(A-pr-qs=x(p-q)\). Divide by \(p-q\): \(x=(A-pr-qs)/(p-q)\).
Question 42
Explanation
Multiply: \(y(cx+d)=ax+b\), so \(ycx+yd=ax+b\). Collect target terms: \((yc-a)x=b-yd\). Divide to get \(x=(b-yd)/(yc-a)\).
Question 43
Explanation
Subtract \(1/R_1\): \(1/R_2=1/R-1/R_1=(R_1-R)/(RR_1)\). Taking reciprocals gives \(R_2=RR_1/(R_1-R)\).
Question 44
Explanation
Multiply: \(P(x+b)=x-a\). Then \(Px+Pb=x-a\), so \((P-1)x=-a-Pb\). Multiplying top and bottom by \(-1\) gives \(x=(a+Pb)/(1-P)\).
Question 45
Explanation
Divide by \(2\pi r\): \(S/(2\pi r)=r+h\). Subtract \(r\): \(h=S/(2\pi r)-r\).
Question 46
Explanation
Factor all three target terms: \(K=x(m+n+p)\). Divide by the complete factored coefficient to get \(x=K/(m+n+p)\).
Question 47
Explanation
Collect target terms: \(ax-cx=d+b\), so \(x(a-c)=d+b\). The correct denominator is \(a-c\); adding coefficients would not represent subtracting \(cx\).
Question 48
Explanation
Cross-multiply: \(d(x-a)=b(x+c)\). Expand: \(dx-da=bx+bc\). Collect: \(x(d-b)=da+bc\). Divide by \(d-b\).
Question 49
Explanation
Factor \(x\): \(T=x[(a+b)-(c-d)]=x(a+b-c+d)\). Divide by the complete coefficient to isolate \(x\).
Question 50
Explanation
Subtract \(r\): \(M-r=px+qx\). Factor \(x(p+q)\), then divide: \(x=(M-r)/(p+q)\).
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