Q. Solve the system of equations by substitution.x=6x+2y−z=18−2x+3y+3z=15
Substitute x into equations: First, we need to substitute x=6 into the second and third equations.x+2y−z=18 becomes 6+2y−z=18.−2x+3y+3z=15 becomes −2(6)+3y+3z=15.
Solve for z: Now, let's solve the first substituted equation for z.6+2y−z=182y−z=18−62y−z=12z=2y−12
Substitute z and x into third equation: Next, we'll substitute z and x into the third equation.−2(6)+3y+3z=15−12+3y+3(2y−12)=15
Simplify the equation: Simplify the equation.−12+3y+6y−36=159y−48=15
Add 48 to both sides: Add 48 to both sides to solve for y.9y−48+48=15+489y=63
Divide by 9 to find y: Divide both sides by 9 to find y.99y=963y=7
Find z using equation: Now we have y, let's find z using the equation z=2y−12. z=2(7)−12 z=14−12 z=2
Check values in original equations: We have x=6, y=7, and z=2. Let's check these values in the original equations.First equation: x=6 (This is given, so it's correct.)Second equation: x+2y−z=18 becomes 6+2(7)−2=18, which simplifies to 6+14−2=18, and that's 18=18, so it's correct.Third equation: −2x+3y+3z=15 becomes −2(6)+3(7)+3(2)=15, which simplifies to y=70, and that's y=71, so it's correct.
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